How to solve $sin x +cos x = 1$?












5














No matter how I do it, I always end up with $x = 0, 90, 270$ and $360$. All of those except $270$ is right, but I can't quite figure out how to get the $270$ degrees out of the answer. I've tried using trig identities, I've tried squaring both sides, but I always end up with $$2sin xcos x$$ which then leads me to $x = 0, 90, 270, 360$. But $$sin (270) + cos (270) = -1$$ so I'm doing something wrong.










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  • 1




    flooding with answers !!!!!!
    – Suraj M S
    Dec 25 '13 at 18:50










  • There are multiple approaches to solving a linear equation in sine and cosine, so that isn't too surprising...
    – colormegone
    Dec 25 '13 at 19:05










  • You have squared, so you get additional solutions. Just throw away those that do not satisfy the original equation.
    – Carsten S
    Dec 25 '13 at 19:24










  • You're not doing anything “wrong”. It's simply that $A^2=B^2$ means $|A|=|B|iff A=pm B$. Obviously, $|-1|=|1|$. All that's left to do now is to trim those “extra” solutions, that's all. :-)
    – Lucian
    Dec 25 '13 at 21:01


















5














No matter how I do it, I always end up with $x = 0, 90, 270$ and $360$. All of those except $270$ is right, but I can't quite figure out how to get the $270$ degrees out of the answer. I've tried using trig identities, I've tried squaring both sides, but I always end up with $$2sin xcos x$$ which then leads me to $x = 0, 90, 270, 360$. But $$sin (270) + cos (270) = -1$$ so I'm doing something wrong.










share|cite|improve this question




















  • 1




    flooding with answers !!!!!!
    – Suraj M S
    Dec 25 '13 at 18:50










  • There are multiple approaches to solving a linear equation in sine and cosine, so that isn't too surprising...
    – colormegone
    Dec 25 '13 at 19:05










  • You have squared, so you get additional solutions. Just throw away those that do not satisfy the original equation.
    – Carsten S
    Dec 25 '13 at 19:24










  • You're not doing anything “wrong”. It's simply that $A^2=B^2$ means $|A|=|B|iff A=pm B$. Obviously, $|-1|=|1|$. All that's left to do now is to trim those “extra” solutions, that's all. :-)
    – Lucian
    Dec 25 '13 at 21:01
















5












5








5


0





No matter how I do it, I always end up with $x = 0, 90, 270$ and $360$. All of those except $270$ is right, but I can't quite figure out how to get the $270$ degrees out of the answer. I've tried using trig identities, I've tried squaring both sides, but I always end up with $$2sin xcos x$$ which then leads me to $x = 0, 90, 270, 360$. But $$sin (270) + cos (270) = -1$$ so I'm doing something wrong.










share|cite|improve this question















No matter how I do it, I always end up with $x = 0, 90, 270$ and $360$. All of those except $270$ is right, but I can't quite figure out how to get the $270$ degrees out of the answer. I've tried using trig identities, I've tried squaring both sides, but I always end up with $$2sin xcos x$$ which then leads me to $x = 0, 90, 270, 360$. But $$sin (270) + cos (270) = -1$$ so I'm doing something wrong.







algebra-precalculus trigonometry






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edited Dec 25 '13 at 18:19









mrs

1




1










asked Dec 25 '13 at 18:16









Threethumb

43631531




43631531








  • 1




    flooding with answers !!!!!!
    – Suraj M S
    Dec 25 '13 at 18:50










  • There are multiple approaches to solving a linear equation in sine and cosine, so that isn't too surprising...
    – colormegone
    Dec 25 '13 at 19:05










  • You have squared, so you get additional solutions. Just throw away those that do not satisfy the original equation.
    – Carsten S
    Dec 25 '13 at 19:24










  • You're not doing anything “wrong”. It's simply that $A^2=B^2$ means $|A|=|B|iff A=pm B$. Obviously, $|-1|=|1|$. All that's left to do now is to trim those “extra” solutions, that's all. :-)
    – Lucian
    Dec 25 '13 at 21:01
















  • 1




    flooding with answers !!!!!!
    – Suraj M S
    Dec 25 '13 at 18:50










  • There are multiple approaches to solving a linear equation in sine and cosine, so that isn't too surprising...
    – colormegone
    Dec 25 '13 at 19:05










  • You have squared, so you get additional solutions. Just throw away those that do not satisfy the original equation.
    – Carsten S
    Dec 25 '13 at 19:24










  • You're not doing anything “wrong”. It's simply that $A^2=B^2$ means $|A|=|B|iff A=pm B$. Obviously, $|-1|=|1|$. All that's left to do now is to trim those “extra” solutions, that's all. :-)
    – Lucian
    Dec 25 '13 at 21:01










1




1




flooding with answers !!!!!!
– Suraj M S
Dec 25 '13 at 18:50




flooding with answers !!!!!!
– Suraj M S
Dec 25 '13 at 18:50












There are multiple approaches to solving a linear equation in sine and cosine, so that isn't too surprising...
– colormegone
Dec 25 '13 at 19:05




There are multiple approaches to solving a linear equation in sine and cosine, so that isn't too surprising...
– colormegone
Dec 25 '13 at 19:05












You have squared, so you get additional solutions. Just throw away those that do not satisfy the original equation.
– Carsten S
Dec 25 '13 at 19:24




You have squared, so you get additional solutions. Just throw away those that do not satisfy the original equation.
– Carsten S
Dec 25 '13 at 19:24












You're not doing anything “wrong”. It's simply that $A^2=B^2$ means $|A|=|B|iff A=pm B$. Obviously, $|-1|=|1|$. All that's left to do now is to trim those “extra” solutions, that's all. :-)
– Lucian
Dec 25 '13 at 21:01






You're not doing anything “wrong”. It's simply that $A^2=B^2$ means $|A|=|B|iff A=pm B$. Obviously, $|-1|=|1|$. All that's left to do now is to trim those “extra” solutions, that's all. :-)
– Lucian
Dec 25 '13 at 21:01












8 Answers
8






active

oldest

votes


















9














The steps you followed are perfectly correct! The only thing you have to do that you haven't done already is check for extraneous solutions. Whenever we square both sides, there is the chance that we get more solutions than we're looking for.



So, you have correctly deduced that the list $0,90,270,360$ (all angles in degrees) contains all potential solutions. After checking, you've noticed that $270$ is not a solution, but the rest are. So, the solutions are $0,90,$ and $360$.



As for why we get extra solutions: notice that although $270$ does not satisfy the original equation, we have
$$
(sin(270)+cos(270))^2=(-1)^2=1
$$
Which makes sense, since we just solved the equation
$$
(sin(x) + cos(x))^2 = 1
$$






share|cite|improve this answer

















  • 2




    This is the one answer that fits the original question.
    – Carsten S
    Dec 25 '13 at 19:28






  • 1




    Okay, I see. But is it mathematically good form to do it this way? Since I'm doing this for an assignment, I'd imagine they want some sort of answer where I arrive at 0, 90 and 360 without needing to check each solution to find a wrong one. Or is it alright to do it by squaring like this and then weeding out erronoeus solutions once I've solved for x?
    – Threethumb
    Dec 25 '13 at 20:21








  • 1




    It is indeed fine mathematical form to do it this way. Yes, you could take an alternate approach not involving extraneous roots (see Fly By Night's answer or RecklessReckoner's), but that doesn't make your method any less valid. I would be very surprised to see a teacher/professor take points off for doing things in the way you've done them.
    – Omnomnomnom
    Dec 25 '13 at 20:44








  • 1




    You might, however, lose points for mentioning $0$ and $360$ as separate solutions, since most contexts call for one or the other. Usually, the question will say something like "find all solutions $x$ (in degrees) to the equation that satisfy $0 leq x < 360$".
    – Omnomnomnom
    Dec 25 '13 at 20:49










  • Ah, they include both 0 and 360 in their own example exercises, so I don't think that will be a problem. Thanks!
    – Threethumb
    Dec 25 '13 at 21:16



















10














Hint:



Use the formula $sin(A+B) equiv sin A cos B + sin B cos A$ to write $sin x + cos x$ in the form $Rsin(x+alpha)$, where $R$ and $alpha$ are numbers that you need to find.



Once you have your $R$ and $alpha$, simply solve $Rsin(x+alpha)=1$.






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    8














    Another way to look at this is to consider the equation as representing the intersection of two "curves" in polar coordinates, one being $ r = 1 $ (the unit circle), the other being the line $ sin theta + cos theta = 1 Rightarrow r sin theta + r cos theta = r Rightarrow r = x + y , $ with $ r $ set equal to 1 . The line has intercepts at ( 1 , 0 ) and ( 0 , 1 ) , meeting the circle at $ theta = 0º text{and} theta = 90º $ . (360º is considered to be merely another "angle-name" for 0º , so it is not really a distinct solution.)






    share|cite|improve this answer





















    • "The line" meaning the line whose equation is $x + y = 1$. This is a beautiful answer.
      – Ryan Reich
      Dec 25 '13 at 19:04










    • Neat solution! I wonder if, were a student to write this solution for this problem in a precalculus class, it could be taken for granted that there are exactly two intersections, or what justification might be necessary.
      – Omnomnomnom
      Dec 25 '13 at 20:55












    • Depending on what topics were covered in their course, the student might solve the system of nonlinear equations, $ x + y = 1 $ and $ x^2 + y^2 = 1 $ ; or solve the improper inequality $ sqrt{1 - x^2} ge 1 - x $ to show that the equality only holds at $ x = 0 , 1 $ and the inequality on $ ( 0 , 1 ) $ , so the curves only make contact at the axis intercepts (and the circle lies "above" the line in between them).
      – colormegone
      Dec 25 '13 at 21:22





















    6














    Use $asin x + bcos x = sqrt{a^2+b^2}cos(x- tan^{-1}(frac ba))$



    Your problem:




    $sqrt2cos(x-fracpi4) = 1$



    $cos(x-fracpi4) = frac1{sqrt2}$



    $x - fracpi4 = 2npi pm fracpi4$




    Solving this gives 0,90,360 as solution.






    share|cite|improve this answer

















    • 8




      This answer would be better if it replaced the unmemorable formula with a method to derive it (as in Fly By Night's answer). Precalculus students already learn too well that math is just about using the right formula.
      – Ryan Reich
      Dec 25 '13 at 19:03










    • Moreover, the formula isn't even really true. It needs to use "atan2" or similar. For instance, replace a and b with their negatives; the LHS changes, but the RHS does not.
      – J Swanson
      Jun 5 '16 at 5:34



















    6














    Hint. Square both sides to get $$(sin x+cos x)^2=1+2sin xcos x=1+sin 2x=1^2=1impliessin 2x=0.$$ The important thing here is to notice $2sin xcos x=sin 2x$. Notice I wrote an implication, not an equivalence, so we can get some extraneous solutions at the end. A routine check can spot all of them.






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    • 1




      Squaring both sides may introduce invalid solutions. For example, if you have $x=1$ and your square both sides you get $x^2=1$, which adds the invalid solution $x=-1$.
      – Fly by Night
      Dec 25 '13 at 18:23










    • @FlybyNight this is an important remark. However, we should always test the solutions, so we are safe.
      – Ian Mateus
      Dec 25 '13 at 18:26










    • I am not keen on this approach. User2369284s solution provides the systematic way of answer this. Avoid squaring whenever possible.
      – Lost1
      Dec 25 '13 at 21:03










    • @Lost1 I don't understand your reluctance. After all, factoring $asin x+bcos x$ gives the same extraneous solutions.
      – Ian Mateus
      Dec 25 '13 at 21:12










    • I dont think it does, pal
      – Lost1
      Dec 25 '13 at 21:58



















    3














    Avoid squaring wherever practicable as it immediately introduces extraneous roots



    Using Weierstrass substitution we have $$frac{2t}{1+t^2}+frac{1-t^2}{1+t^2}=1$$ where $t=tanfrac x2$



    $$implies 2t+1-t^2=1+t^2iff t^2-t=0iff t=1,0$$



    If $displaystyletanfrac x2=0iff frac x2=n180^circiff x=n360^circ$ where $n$ is any integer



    If $displaystyletanfrac x2=1iff frac x2=m180^circ+45^circiff x=m360^circ+90^circ$ where $m$ is any integer






    share|cite|improve this answer





















    • @user107827, Try this method when $$sin x+cos x=-1$$
      – lab bhattacharjee
      Dec 26 '13 at 13:46



















    2














    Using Double-Angle Formulas,



    $$sin x+cos x=1implies 2sinfrac x2cosfrac x2=1-cos x=2sin^2frac x2$$



    $$implies sinfrac x2left(cosfrac x2-sinfrac x2right)=0$$



    $$(i)sinfrac x2=0implies frac x2=npitext{ where }n text{ is any integer}$$



    $$(ii) cosfrac x2-sinfrac x2=0impliescosfrac x2=sinfrac x2iff tanfrac x2=1$$ Find the rest in my other answer






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      1














      $sin x + cos x = 1 rightarrow (sin x + cos x )^2=1^2 iff color{green}{underbrace{(sin^2 x + cos^2 x)}_{=1}}+2sin x cos x =color{green}{1} iff sin x cos x = 0 $



      Now look at the unit circle to see when this is true. After you have found the values of $x$ for which this holds, be sure to check the outcome of $sin x + cos x$ for each of these values, to eliminate the outcomes that generate value $-1$ (these values of $x$ came in because of the squaring - but are not truly solutions).






      share|cite|improve this answer























      • @DanielFischer My bad. Good point.
        – dreamer
        Dec 25 '13 at 18:27










      • Note that squaring may add ghost solutions. You need to mention this. If we start with $x=1$ then there is one solution: $x=1$. Squaring both sides gives $x^2=1$, which has two solutions: $x=pm 1$. One solution, $x=1$, is a correct solution of the original problem while the other, $x=-1$, is a ghost solution introduced by squaring.
        – Fly by Night
        Dec 25 '13 at 19:17












      • @FlybyNight You're right. Therefore, it would be good indeed to check the answer.
        – dreamer
        Dec 25 '13 at 19:22










      • The equation $cos x sin x =0$ has $x=180^{circ}$ and $x=270^{circ}$ as solutions. However $sin(180^{circ})+cos(180^{circ})=-1$ and $sin(270^{circ})+cos(270^{circ})=-1$. The OP asked for solutions to $sin x + cos x =1$.
        – Fly by Night
        Dec 25 '13 at 19:46










      • @FlybyNight No, those are not the only ones. $x=90^{circ}$ and $x=0^{circ}$ are also solutions , which both result in $sin x + cos x =1$.
        – dreamer
        Dec 25 '13 at 20:10











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      8 Answers
      8






      active

      oldest

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      8 Answers
      8






      active

      oldest

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      active

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      active

      oldest

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      9














      The steps you followed are perfectly correct! The only thing you have to do that you haven't done already is check for extraneous solutions. Whenever we square both sides, there is the chance that we get more solutions than we're looking for.



      So, you have correctly deduced that the list $0,90,270,360$ (all angles in degrees) contains all potential solutions. After checking, you've noticed that $270$ is not a solution, but the rest are. So, the solutions are $0,90,$ and $360$.



      As for why we get extra solutions: notice that although $270$ does not satisfy the original equation, we have
      $$
      (sin(270)+cos(270))^2=(-1)^2=1
      $$
      Which makes sense, since we just solved the equation
      $$
      (sin(x) + cos(x))^2 = 1
      $$






      share|cite|improve this answer

















      • 2




        This is the one answer that fits the original question.
        – Carsten S
        Dec 25 '13 at 19:28






      • 1




        Okay, I see. But is it mathematically good form to do it this way? Since I'm doing this for an assignment, I'd imagine they want some sort of answer where I arrive at 0, 90 and 360 without needing to check each solution to find a wrong one. Or is it alright to do it by squaring like this and then weeding out erronoeus solutions once I've solved for x?
        – Threethumb
        Dec 25 '13 at 20:21








      • 1




        It is indeed fine mathematical form to do it this way. Yes, you could take an alternate approach not involving extraneous roots (see Fly By Night's answer or RecklessReckoner's), but that doesn't make your method any less valid. I would be very surprised to see a teacher/professor take points off for doing things in the way you've done them.
        – Omnomnomnom
        Dec 25 '13 at 20:44








      • 1




        You might, however, lose points for mentioning $0$ and $360$ as separate solutions, since most contexts call for one or the other. Usually, the question will say something like "find all solutions $x$ (in degrees) to the equation that satisfy $0 leq x < 360$".
        – Omnomnomnom
        Dec 25 '13 at 20:49










      • Ah, they include both 0 and 360 in their own example exercises, so I don't think that will be a problem. Thanks!
        – Threethumb
        Dec 25 '13 at 21:16
















      9














      The steps you followed are perfectly correct! The only thing you have to do that you haven't done already is check for extraneous solutions. Whenever we square both sides, there is the chance that we get more solutions than we're looking for.



      So, you have correctly deduced that the list $0,90,270,360$ (all angles in degrees) contains all potential solutions. After checking, you've noticed that $270$ is not a solution, but the rest are. So, the solutions are $0,90,$ and $360$.



      As for why we get extra solutions: notice that although $270$ does not satisfy the original equation, we have
      $$
      (sin(270)+cos(270))^2=(-1)^2=1
      $$
      Which makes sense, since we just solved the equation
      $$
      (sin(x) + cos(x))^2 = 1
      $$






      share|cite|improve this answer

















      • 2




        This is the one answer that fits the original question.
        – Carsten S
        Dec 25 '13 at 19:28






      • 1




        Okay, I see. But is it mathematically good form to do it this way? Since I'm doing this for an assignment, I'd imagine they want some sort of answer where I arrive at 0, 90 and 360 without needing to check each solution to find a wrong one. Or is it alright to do it by squaring like this and then weeding out erronoeus solutions once I've solved for x?
        – Threethumb
        Dec 25 '13 at 20:21








      • 1




        It is indeed fine mathematical form to do it this way. Yes, you could take an alternate approach not involving extraneous roots (see Fly By Night's answer or RecklessReckoner's), but that doesn't make your method any less valid. I would be very surprised to see a teacher/professor take points off for doing things in the way you've done them.
        – Omnomnomnom
        Dec 25 '13 at 20:44








      • 1




        You might, however, lose points for mentioning $0$ and $360$ as separate solutions, since most contexts call for one or the other. Usually, the question will say something like "find all solutions $x$ (in degrees) to the equation that satisfy $0 leq x < 360$".
        – Omnomnomnom
        Dec 25 '13 at 20:49










      • Ah, they include both 0 and 360 in their own example exercises, so I don't think that will be a problem. Thanks!
        – Threethumb
        Dec 25 '13 at 21:16














      9












      9








      9






      The steps you followed are perfectly correct! The only thing you have to do that you haven't done already is check for extraneous solutions. Whenever we square both sides, there is the chance that we get more solutions than we're looking for.



      So, you have correctly deduced that the list $0,90,270,360$ (all angles in degrees) contains all potential solutions. After checking, you've noticed that $270$ is not a solution, but the rest are. So, the solutions are $0,90,$ and $360$.



      As for why we get extra solutions: notice that although $270$ does not satisfy the original equation, we have
      $$
      (sin(270)+cos(270))^2=(-1)^2=1
      $$
      Which makes sense, since we just solved the equation
      $$
      (sin(x) + cos(x))^2 = 1
      $$






      share|cite|improve this answer












      The steps you followed are perfectly correct! The only thing you have to do that you haven't done already is check for extraneous solutions. Whenever we square both sides, there is the chance that we get more solutions than we're looking for.



      So, you have correctly deduced that the list $0,90,270,360$ (all angles in degrees) contains all potential solutions. After checking, you've noticed that $270$ is not a solution, but the rest are. So, the solutions are $0,90,$ and $360$.



      As for why we get extra solutions: notice that although $270$ does not satisfy the original equation, we have
      $$
      (sin(270)+cos(270))^2=(-1)^2=1
      $$
      Which makes sense, since we just solved the equation
      $$
      (sin(x) + cos(x))^2 = 1
      $$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 25 '13 at 18:42









      Omnomnomnom

      126k788176




      126k788176








      • 2




        This is the one answer that fits the original question.
        – Carsten S
        Dec 25 '13 at 19:28






      • 1




        Okay, I see. But is it mathematically good form to do it this way? Since I'm doing this for an assignment, I'd imagine they want some sort of answer where I arrive at 0, 90 and 360 without needing to check each solution to find a wrong one. Or is it alright to do it by squaring like this and then weeding out erronoeus solutions once I've solved for x?
        – Threethumb
        Dec 25 '13 at 20:21








      • 1




        It is indeed fine mathematical form to do it this way. Yes, you could take an alternate approach not involving extraneous roots (see Fly By Night's answer or RecklessReckoner's), but that doesn't make your method any less valid. I would be very surprised to see a teacher/professor take points off for doing things in the way you've done them.
        – Omnomnomnom
        Dec 25 '13 at 20:44








      • 1




        You might, however, lose points for mentioning $0$ and $360$ as separate solutions, since most contexts call for one or the other. Usually, the question will say something like "find all solutions $x$ (in degrees) to the equation that satisfy $0 leq x < 360$".
        – Omnomnomnom
        Dec 25 '13 at 20:49










      • Ah, they include both 0 and 360 in their own example exercises, so I don't think that will be a problem. Thanks!
        – Threethumb
        Dec 25 '13 at 21:16














      • 2




        This is the one answer that fits the original question.
        – Carsten S
        Dec 25 '13 at 19:28






      • 1




        Okay, I see. But is it mathematically good form to do it this way? Since I'm doing this for an assignment, I'd imagine they want some sort of answer where I arrive at 0, 90 and 360 without needing to check each solution to find a wrong one. Or is it alright to do it by squaring like this and then weeding out erronoeus solutions once I've solved for x?
        – Threethumb
        Dec 25 '13 at 20:21








      • 1




        It is indeed fine mathematical form to do it this way. Yes, you could take an alternate approach not involving extraneous roots (see Fly By Night's answer or RecklessReckoner's), but that doesn't make your method any less valid. I would be very surprised to see a teacher/professor take points off for doing things in the way you've done them.
        – Omnomnomnom
        Dec 25 '13 at 20:44








      • 1




        You might, however, lose points for mentioning $0$ and $360$ as separate solutions, since most contexts call for one or the other. Usually, the question will say something like "find all solutions $x$ (in degrees) to the equation that satisfy $0 leq x < 360$".
        – Omnomnomnom
        Dec 25 '13 at 20:49










      • Ah, they include both 0 and 360 in their own example exercises, so I don't think that will be a problem. Thanks!
        – Threethumb
        Dec 25 '13 at 21:16








      2




      2




      This is the one answer that fits the original question.
      – Carsten S
      Dec 25 '13 at 19:28




      This is the one answer that fits the original question.
      – Carsten S
      Dec 25 '13 at 19:28




      1




      1




      Okay, I see. But is it mathematically good form to do it this way? Since I'm doing this for an assignment, I'd imagine they want some sort of answer where I arrive at 0, 90 and 360 without needing to check each solution to find a wrong one. Or is it alright to do it by squaring like this and then weeding out erronoeus solutions once I've solved for x?
      – Threethumb
      Dec 25 '13 at 20:21






      Okay, I see. But is it mathematically good form to do it this way? Since I'm doing this for an assignment, I'd imagine they want some sort of answer where I arrive at 0, 90 and 360 without needing to check each solution to find a wrong one. Or is it alright to do it by squaring like this and then weeding out erronoeus solutions once I've solved for x?
      – Threethumb
      Dec 25 '13 at 20:21






      1




      1




      It is indeed fine mathematical form to do it this way. Yes, you could take an alternate approach not involving extraneous roots (see Fly By Night's answer or RecklessReckoner's), but that doesn't make your method any less valid. I would be very surprised to see a teacher/professor take points off for doing things in the way you've done them.
      – Omnomnomnom
      Dec 25 '13 at 20:44






      It is indeed fine mathematical form to do it this way. Yes, you could take an alternate approach not involving extraneous roots (see Fly By Night's answer or RecklessReckoner's), but that doesn't make your method any less valid. I would be very surprised to see a teacher/professor take points off for doing things in the way you've done them.
      – Omnomnomnom
      Dec 25 '13 at 20:44






      1




      1




      You might, however, lose points for mentioning $0$ and $360$ as separate solutions, since most contexts call for one or the other. Usually, the question will say something like "find all solutions $x$ (in degrees) to the equation that satisfy $0 leq x < 360$".
      – Omnomnomnom
      Dec 25 '13 at 20:49




      You might, however, lose points for mentioning $0$ and $360$ as separate solutions, since most contexts call for one or the other. Usually, the question will say something like "find all solutions $x$ (in degrees) to the equation that satisfy $0 leq x < 360$".
      – Omnomnomnom
      Dec 25 '13 at 20:49












      Ah, they include both 0 and 360 in their own example exercises, so I don't think that will be a problem. Thanks!
      – Threethumb
      Dec 25 '13 at 21:16




      Ah, they include both 0 and 360 in their own example exercises, so I don't think that will be a problem. Thanks!
      – Threethumb
      Dec 25 '13 at 21:16











      10














      Hint:



      Use the formula $sin(A+B) equiv sin A cos B + sin B cos A$ to write $sin x + cos x$ in the form $Rsin(x+alpha)$, where $R$ and $alpha$ are numbers that you need to find.



      Once you have your $R$ and $alpha$, simply solve $Rsin(x+alpha)=1$.






      share|cite|improve this answer


























        10














        Hint:



        Use the formula $sin(A+B) equiv sin A cos B + sin B cos A$ to write $sin x + cos x$ in the form $Rsin(x+alpha)$, where $R$ and $alpha$ are numbers that you need to find.



        Once you have your $R$ and $alpha$, simply solve $Rsin(x+alpha)=1$.






        share|cite|improve this answer
























          10












          10








          10






          Hint:



          Use the formula $sin(A+B) equiv sin A cos B + sin B cos A$ to write $sin x + cos x$ in the form $Rsin(x+alpha)$, where $R$ and $alpha$ are numbers that you need to find.



          Once you have your $R$ and $alpha$, simply solve $Rsin(x+alpha)=1$.






          share|cite|improve this answer












          Hint:



          Use the formula $sin(A+B) equiv sin A cos B + sin B cos A$ to write $sin x + cos x$ in the form $Rsin(x+alpha)$, where $R$ and $alpha$ are numbers that you need to find.



          Once you have your $R$ and $alpha$, simply solve $Rsin(x+alpha)=1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 25 '13 at 18:17









          Fly by Night

          25.7k32978




          25.7k32978























              8














              Another way to look at this is to consider the equation as representing the intersection of two "curves" in polar coordinates, one being $ r = 1 $ (the unit circle), the other being the line $ sin theta + cos theta = 1 Rightarrow r sin theta + r cos theta = r Rightarrow r = x + y , $ with $ r $ set equal to 1 . The line has intercepts at ( 1 , 0 ) and ( 0 , 1 ) , meeting the circle at $ theta = 0º text{and} theta = 90º $ . (360º is considered to be merely another "angle-name" for 0º , so it is not really a distinct solution.)






              share|cite|improve this answer





















              • "The line" meaning the line whose equation is $x + y = 1$. This is a beautiful answer.
                – Ryan Reich
                Dec 25 '13 at 19:04










              • Neat solution! I wonder if, were a student to write this solution for this problem in a precalculus class, it could be taken for granted that there are exactly two intersections, or what justification might be necessary.
                – Omnomnomnom
                Dec 25 '13 at 20:55












              • Depending on what topics were covered in their course, the student might solve the system of nonlinear equations, $ x + y = 1 $ and $ x^2 + y^2 = 1 $ ; or solve the improper inequality $ sqrt{1 - x^2} ge 1 - x $ to show that the equality only holds at $ x = 0 , 1 $ and the inequality on $ ( 0 , 1 ) $ , so the curves only make contact at the axis intercepts (and the circle lies "above" the line in between them).
                – colormegone
                Dec 25 '13 at 21:22


















              8














              Another way to look at this is to consider the equation as representing the intersection of two "curves" in polar coordinates, one being $ r = 1 $ (the unit circle), the other being the line $ sin theta + cos theta = 1 Rightarrow r sin theta + r cos theta = r Rightarrow r = x + y , $ with $ r $ set equal to 1 . The line has intercepts at ( 1 , 0 ) and ( 0 , 1 ) , meeting the circle at $ theta = 0º text{and} theta = 90º $ . (360º is considered to be merely another "angle-name" for 0º , so it is not really a distinct solution.)






              share|cite|improve this answer





















              • "The line" meaning the line whose equation is $x + y = 1$. This is a beautiful answer.
                – Ryan Reich
                Dec 25 '13 at 19:04










              • Neat solution! I wonder if, were a student to write this solution for this problem in a precalculus class, it could be taken for granted that there are exactly two intersections, or what justification might be necessary.
                – Omnomnomnom
                Dec 25 '13 at 20:55












              • Depending on what topics were covered in their course, the student might solve the system of nonlinear equations, $ x + y = 1 $ and $ x^2 + y^2 = 1 $ ; or solve the improper inequality $ sqrt{1 - x^2} ge 1 - x $ to show that the equality only holds at $ x = 0 , 1 $ and the inequality on $ ( 0 , 1 ) $ , so the curves only make contact at the axis intercepts (and the circle lies "above" the line in between them).
                – colormegone
                Dec 25 '13 at 21:22
















              8












              8








              8






              Another way to look at this is to consider the equation as representing the intersection of two "curves" in polar coordinates, one being $ r = 1 $ (the unit circle), the other being the line $ sin theta + cos theta = 1 Rightarrow r sin theta + r cos theta = r Rightarrow r = x + y , $ with $ r $ set equal to 1 . The line has intercepts at ( 1 , 0 ) and ( 0 , 1 ) , meeting the circle at $ theta = 0º text{and} theta = 90º $ . (360º is considered to be merely another "angle-name" for 0º , so it is not really a distinct solution.)






              share|cite|improve this answer












              Another way to look at this is to consider the equation as representing the intersection of two "curves" in polar coordinates, one being $ r = 1 $ (the unit circle), the other being the line $ sin theta + cos theta = 1 Rightarrow r sin theta + r cos theta = r Rightarrow r = x + y , $ with $ r $ set equal to 1 . The line has intercepts at ( 1 , 0 ) and ( 0 , 1 ) , meeting the circle at $ theta = 0º text{and} theta = 90º $ . (360º is considered to be merely another "angle-name" for 0º , so it is not really a distinct solution.)







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 25 '13 at 19:02









              colormegone

              9,62321342




              9,62321342












              • "The line" meaning the line whose equation is $x + y = 1$. This is a beautiful answer.
                – Ryan Reich
                Dec 25 '13 at 19:04










              • Neat solution! I wonder if, were a student to write this solution for this problem in a precalculus class, it could be taken for granted that there are exactly two intersections, or what justification might be necessary.
                – Omnomnomnom
                Dec 25 '13 at 20:55












              • Depending on what topics were covered in their course, the student might solve the system of nonlinear equations, $ x + y = 1 $ and $ x^2 + y^2 = 1 $ ; or solve the improper inequality $ sqrt{1 - x^2} ge 1 - x $ to show that the equality only holds at $ x = 0 , 1 $ and the inequality on $ ( 0 , 1 ) $ , so the curves only make contact at the axis intercepts (and the circle lies "above" the line in between them).
                – colormegone
                Dec 25 '13 at 21:22




















              • "The line" meaning the line whose equation is $x + y = 1$. This is a beautiful answer.
                – Ryan Reich
                Dec 25 '13 at 19:04










              • Neat solution! I wonder if, were a student to write this solution for this problem in a precalculus class, it could be taken for granted that there are exactly two intersections, or what justification might be necessary.
                – Omnomnomnom
                Dec 25 '13 at 20:55












              • Depending on what topics were covered in their course, the student might solve the system of nonlinear equations, $ x + y = 1 $ and $ x^2 + y^2 = 1 $ ; or solve the improper inequality $ sqrt{1 - x^2} ge 1 - x $ to show that the equality only holds at $ x = 0 , 1 $ and the inequality on $ ( 0 , 1 ) $ , so the curves only make contact at the axis intercepts (and the circle lies "above" the line in between them).
                – colormegone
                Dec 25 '13 at 21:22


















              "The line" meaning the line whose equation is $x + y = 1$. This is a beautiful answer.
              – Ryan Reich
              Dec 25 '13 at 19:04




              "The line" meaning the line whose equation is $x + y = 1$. This is a beautiful answer.
              – Ryan Reich
              Dec 25 '13 at 19:04












              Neat solution! I wonder if, were a student to write this solution for this problem in a precalculus class, it could be taken for granted that there are exactly two intersections, or what justification might be necessary.
              – Omnomnomnom
              Dec 25 '13 at 20:55






              Neat solution! I wonder if, were a student to write this solution for this problem in a precalculus class, it could be taken for granted that there are exactly two intersections, or what justification might be necessary.
              – Omnomnomnom
              Dec 25 '13 at 20:55














              Depending on what topics were covered in their course, the student might solve the system of nonlinear equations, $ x + y = 1 $ and $ x^2 + y^2 = 1 $ ; or solve the improper inequality $ sqrt{1 - x^2} ge 1 - x $ to show that the equality only holds at $ x = 0 , 1 $ and the inequality on $ ( 0 , 1 ) $ , so the curves only make contact at the axis intercepts (and the circle lies "above" the line in between them).
              – colormegone
              Dec 25 '13 at 21:22






              Depending on what topics were covered in their course, the student might solve the system of nonlinear equations, $ x + y = 1 $ and $ x^2 + y^2 = 1 $ ; or solve the improper inequality $ sqrt{1 - x^2} ge 1 - x $ to show that the equality only holds at $ x = 0 , 1 $ and the inequality on $ ( 0 , 1 ) $ , so the curves only make contact at the axis intercepts (and the circle lies "above" the line in between them).
              – colormegone
              Dec 25 '13 at 21:22













              6














              Use $asin x + bcos x = sqrt{a^2+b^2}cos(x- tan^{-1}(frac ba))$



              Your problem:




              $sqrt2cos(x-fracpi4) = 1$



              $cos(x-fracpi4) = frac1{sqrt2}$



              $x - fracpi4 = 2npi pm fracpi4$




              Solving this gives 0,90,360 as solution.






              share|cite|improve this answer

















              • 8




                This answer would be better if it replaced the unmemorable formula with a method to derive it (as in Fly By Night's answer). Precalculus students already learn too well that math is just about using the right formula.
                – Ryan Reich
                Dec 25 '13 at 19:03










              • Moreover, the formula isn't even really true. It needs to use "atan2" or similar. For instance, replace a and b with their negatives; the LHS changes, but the RHS does not.
                – J Swanson
                Jun 5 '16 at 5:34
















              6














              Use $asin x + bcos x = sqrt{a^2+b^2}cos(x- tan^{-1}(frac ba))$



              Your problem:




              $sqrt2cos(x-fracpi4) = 1$



              $cos(x-fracpi4) = frac1{sqrt2}$



              $x - fracpi4 = 2npi pm fracpi4$




              Solving this gives 0,90,360 as solution.






              share|cite|improve this answer

















              • 8




                This answer would be better if it replaced the unmemorable formula with a method to derive it (as in Fly By Night's answer). Precalculus students already learn too well that math is just about using the right formula.
                – Ryan Reich
                Dec 25 '13 at 19:03










              • Moreover, the formula isn't even really true. It needs to use "atan2" or similar. For instance, replace a and b with their negatives; the LHS changes, but the RHS does not.
                – J Swanson
                Jun 5 '16 at 5:34














              6












              6








              6






              Use $asin x + bcos x = sqrt{a^2+b^2}cos(x- tan^{-1}(frac ba))$



              Your problem:




              $sqrt2cos(x-fracpi4) = 1$



              $cos(x-fracpi4) = frac1{sqrt2}$



              $x - fracpi4 = 2npi pm fracpi4$




              Solving this gives 0,90,360 as solution.






              share|cite|improve this answer












              Use $asin x + bcos x = sqrt{a^2+b^2}cos(x- tan^{-1}(frac ba))$



              Your problem:




              $sqrt2cos(x-fracpi4) = 1$



              $cos(x-fracpi4) = frac1{sqrt2}$



              $x - fracpi4 = 2npi pm fracpi4$




              Solving this gives 0,90,360 as solution.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 25 '13 at 18:31









              user2369284

              1,393827




              1,393827








              • 8




                This answer would be better if it replaced the unmemorable formula with a method to derive it (as in Fly By Night's answer). Precalculus students already learn too well that math is just about using the right formula.
                – Ryan Reich
                Dec 25 '13 at 19:03










              • Moreover, the formula isn't even really true. It needs to use "atan2" or similar. For instance, replace a and b with their negatives; the LHS changes, but the RHS does not.
                – J Swanson
                Jun 5 '16 at 5:34














              • 8




                This answer would be better if it replaced the unmemorable formula with a method to derive it (as in Fly By Night's answer). Precalculus students already learn too well that math is just about using the right formula.
                – Ryan Reich
                Dec 25 '13 at 19:03










              • Moreover, the formula isn't even really true. It needs to use "atan2" or similar. For instance, replace a and b with their negatives; the LHS changes, but the RHS does not.
                – J Swanson
                Jun 5 '16 at 5:34








              8




              8




              This answer would be better if it replaced the unmemorable formula with a method to derive it (as in Fly By Night's answer). Precalculus students already learn too well that math is just about using the right formula.
              – Ryan Reich
              Dec 25 '13 at 19:03




              This answer would be better if it replaced the unmemorable formula with a method to derive it (as in Fly By Night's answer). Precalculus students already learn too well that math is just about using the right formula.
              – Ryan Reich
              Dec 25 '13 at 19:03












              Moreover, the formula isn't even really true. It needs to use "atan2" or similar. For instance, replace a and b with their negatives; the LHS changes, but the RHS does not.
              – J Swanson
              Jun 5 '16 at 5:34




              Moreover, the formula isn't even really true. It needs to use "atan2" or similar. For instance, replace a and b with their negatives; the LHS changes, but the RHS does not.
              – J Swanson
              Jun 5 '16 at 5:34











              6














              Hint. Square both sides to get $$(sin x+cos x)^2=1+2sin xcos x=1+sin 2x=1^2=1impliessin 2x=0.$$ The important thing here is to notice $2sin xcos x=sin 2x$. Notice I wrote an implication, not an equivalence, so we can get some extraneous solutions at the end. A routine check can spot all of them.






              share|cite|improve this answer



















              • 1




                Squaring both sides may introduce invalid solutions. For example, if you have $x=1$ and your square both sides you get $x^2=1$, which adds the invalid solution $x=-1$.
                – Fly by Night
                Dec 25 '13 at 18:23










              • @FlybyNight this is an important remark. However, we should always test the solutions, so we are safe.
                – Ian Mateus
                Dec 25 '13 at 18:26










              • I am not keen on this approach. User2369284s solution provides the systematic way of answer this. Avoid squaring whenever possible.
                – Lost1
                Dec 25 '13 at 21:03










              • @Lost1 I don't understand your reluctance. After all, factoring $asin x+bcos x$ gives the same extraneous solutions.
                – Ian Mateus
                Dec 25 '13 at 21:12










              • I dont think it does, pal
                – Lost1
                Dec 25 '13 at 21:58
















              6














              Hint. Square both sides to get $$(sin x+cos x)^2=1+2sin xcos x=1+sin 2x=1^2=1impliessin 2x=0.$$ The important thing here is to notice $2sin xcos x=sin 2x$. Notice I wrote an implication, not an equivalence, so we can get some extraneous solutions at the end. A routine check can spot all of them.






              share|cite|improve this answer



















              • 1




                Squaring both sides may introduce invalid solutions. For example, if you have $x=1$ and your square both sides you get $x^2=1$, which adds the invalid solution $x=-1$.
                – Fly by Night
                Dec 25 '13 at 18:23










              • @FlybyNight this is an important remark. However, we should always test the solutions, so we are safe.
                – Ian Mateus
                Dec 25 '13 at 18:26










              • I am not keen on this approach. User2369284s solution provides the systematic way of answer this. Avoid squaring whenever possible.
                – Lost1
                Dec 25 '13 at 21:03










              • @Lost1 I don't understand your reluctance. After all, factoring $asin x+bcos x$ gives the same extraneous solutions.
                – Ian Mateus
                Dec 25 '13 at 21:12










              • I dont think it does, pal
                – Lost1
                Dec 25 '13 at 21:58














              6












              6








              6






              Hint. Square both sides to get $$(sin x+cos x)^2=1+2sin xcos x=1+sin 2x=1^2=1impliessin 2x=0.$$ The important thing here is to notice $2sin xcos x=sin 2x$. Notice I wrote an implication, not an equivalence, so we can get some extraneous solutions at the end. A routine check can spot all of them.






              share|cite|improve this answer














              Hint. Square both sides to get $$(sin x+cos x)^2=1+2sin xcos x=1+sin 2x=1^2=1impliessin 2x=0.$$ The important thing here is to notice $2sin xcos x=sin 2x$. Notice I wrote an implication, not an equivalence, so we can get some extraneous solutions at the end. A routine check can spot all of them.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Dec 25 '13 at 18:31

























              answered Dec 25 '13 at 18:22









              Ian Mateus

              4,66032452




              4,66032452








              • 1




                Squaring both sides may introduce invalid solutions. For example, if you have $x=1$ and your square both sides you get $x^2=1$, which adds the invalid solution $x=-1$.
                – Fly by Night
                Dec 25 '13 at 18:23










              • @FlybyNight this is an important remark. However, we should always test the solutions, so we are safe.
                – Ian Mateus
                Dec 25 '13 at 18:26










              • I am not keen on this approach. User2369284s solution provides the systematic way of answer this. Avoid squaring whenever possible.
                – Lost1
                Dec 25 '13 at 21:03










              • @Lost1 I don't understand your reluctance. After all, factoring $asin x+bcos x$ gives the same extraneous solutions.
                – Ian Mateus
                Dec 25 '13 at 21:12










              • I dont think it does, pal
                – Lost1
                Dec 25 '13 at 21:58














              • 1




                Squaring both sides may introduce invalid solutions. For example, if you have $x=1$ and your square both sides you get $x^2=1$, which adds the invalid solution $x=-1$.
                – Fly by Night
                Dec 25 '13 at 18:23










              • @FlybyNight this is an important remark. However, we should always test the solutions, so we are safe.
                – Ian Mateus
                Dec 25 '13 at 18:26










              • I am not keen on this approach. User2369284s solution provides the systematic way of answer this. Avoid squaring whenever possible.
                – Lost1
                Dec 25 '13 at 21:03










              • @Lost1 I don't understand your reluctance. After all, factoring $asin x+bcos x$ gives the same extraneous solutions.
                – Ian Mateus
                Dec 25 '13 at 21:12










              • I dont think it does, pal
                – Lost1
                Dec 25 '13 at 21:58








              1




              1




              Squaring both sides may introduce invalid solutions. For example, if you have $x=1$ and your square both sides you get $x^2=1$, which adds the invalid solution $x=-1$.
              – Fly by Night
              Dec 25 '13 at 18:23




              Squaring both sides may introduce invalid solutions. For example, if you have $x=1$ and your square both sides you get $x^2=1$, which adds the invalid solution $x=-1$.
              – Fly by Night
              Dec 25 '13 at 18:23












              @FlybyNight this is an important remark. However, we should always test the solutions, so we are safe.
              – Ian Mateus
              Dec 25 '13 at 18:26




              @FlybyNight this is an important remark. However, we should always test the solutions, so we are safe.
              – Ian Mateus
              Dec 25 '13 at 18:26












              I am not keen on this approach. User2369284s solution provides the systematic way of answer this. Avoid squaring whenever possible.
              – Lost1
              Dec 25 '13 at 21:03




              I am not keen on this approach. User2369284s solution provides the systematic way of answer this. Avoid squaring whenever possible.
              – Lost1
              Dec 25 '13 at 21:03












              @Lost1 I don't understand your reluctance. After all, factoring $asin x+bcos x$ gives the same extraneous solutions.
              – Ian Mateus
              Dec 25 '13 at 21:12




              @Lost1 I don't understand your reluctance. After all, factoring $asin x+bcos x$ gives the same extraneous solutions.
              – Ian Mateus
              Dec 25 '13 at 21:12












              I dont think it does, pal
              – Lost1
              Dec 25 '13 at 21:58




              I dont think it does, pal
              – Lost1
              Dec 25 '13 at 21:58











              3














              Avoid squaring wherever practicable as it immediately introduces extraneous roots



              Using Weierstrass substitution we have $$frac{2t}{1+t^2}+frac{1-t^2}{1+t^2}=1$$ where $t=tanfrac x2$



              $$implies 2t+1-t^2=1+t^2iff t^2-t=0iff t=1,0$$



              If $displaystyletanfrac x2=0iff frac x2=n180^circiff x=n360^circ$ where $n$ is any integer



              If $displaystyletanfrac x2=1iff frac x2=m180^circ+45^circiff x=m360^circ+90^circ$ where $m$ is any integer






              share|cite|improve this answer





















              • @user107827, Try this method when $$sin x+cos x=-1$$
                – lab bhattacharjee
                Dec 26 '13 at 13:46
















              3














              Avoid squaring wherever practicable as it immediately introduces extraneous roots



              Using Weierstrass substitution we have $$frac{2t}{1+t^2}+frac{1-t^2}{1+t^2}=1$$ where $t=tanfrac x2$



              $$implies 2t+1-t^2=1+t^2iff t^2-t=0iff t=1,0$$



              If $displaystyletanfrac x2=0iff frac x2=n180^circiff x=n360^circ$ where $n$ is any integer



              If $displaystyletanfrac x2=1iff frac x2=m180^circ+45^circiff x=m360^circ+90^circ$ where $m$ is any integer






              share|cite|improve this answer





















              • @user107827, Try this method when $$sin x+cos x=-1$$
                – lab bhattacharjee
                Dec 26 '13 at 13:46














              3












              3








              3






              Avoid squaring wherever practicable as it immediately introduces extraneous roots



              Using Weierstrass substitution we have $$frac{2t}{1+t^2}+frac{1-t^2}{1+t^2}=1$$ where $t=tanfrac x2$



              $$implies 2t+1-t^2=1+t^2iff t^2-t=0iff t=1,0$$



              If $displaystyletanfrac x2=0iff frac x2=n180^circiff x=n360^circ$ where $n$ is any integer



              If $displaystyletanfrac x2=1iff frac x2=m180^circ+45^circiff x=m360^circ+90^circ$ where $m$ is any integer






              share|cite|improve this answer












              Avoid squaring wherever practicable as it immediately introduces extraneous roots



              Using Weierstrass substitution we have $$frac{2t}{1+t^2}+frac{1-t^2}{1+t^2}=1$$ where $t=tanfrac x2$



              $$implies 2t+1-t^2=1+t^2iff t^2-t=0iff t=1,0$$



              If $displaystyletanfrac x2=0iff frac x2=n180^circiff x=n360^circ$ where $n$ is any integer



              If $displaystyletanfrac x2=1iff frac x2=m180^circ+45^circiff x=m360^circ+90^circ$ where $m$ is any integer







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 26 '13 at 3:23









              lab bhattacharjee

              223k15156274




              223k15156274












              • @user107827, Try this method when $$sin x+cos x=-1$$
                – lab bhattacharjee
                Dec 26 '13 at 13:46


















              • @user107827, Try this method when $$sin x+cos x=-1$$
                – lab bhattacharjee
                Dec 26 '13 at 13:46
















              @user107827, Try this method when $$sin x+cos x=-1$$
              – lab bhattacharjee
              Dec 26 '13 at 13:46




              @user107827, Try this method when $$sin x+cos x=-1$$
              – lab bhattacharjee
              Dec 26 '13 at 13:46











              2














              Using Double-Angle Formulas,



              $$sin x+cos x=1implies 2sinfrac x2cosfrac x2=1-cos x=2sin^2frac x2$$



              $$implies sinfrac x2left(cosfrac x2-sinfrac x2right)=0$$



              $$(i)sinfrac x2=0implies frac x2=npitext{ where }n text{ is any integer}$$



              $$(ii) cosfrac x2-sinfrac x2=0impliescosfrac x2=sinfrac x2iff tanfrac x2=1$$ Find the rest in my other answer






              share|cite|improve this answer


























                2














                Using Double-Angle Formulas,



                $$sin x+cos x=1implies 2sinfrac x2cosfrac x2=1-cos x=2sin^2frac x2$$



                $$implies sinfrac x2left(cosfrac x2-sinfrac x2right)=0$$



                $$(i)sinfrac x2=0implies frac x2=npitext{ where }n text{ is any integer}$$



                $$(ii) cosfrac x2-sinfrac x2=0impliescosfrac x2=sinfrac x2iff tanfrac x2=1$$ Find the rest in my other answer






                share|cite|improve this answer
























                  2












                  2








                  2






                  Using Double-Angle Formulas,



                  $$sin x+cos x=1implies 2sinfrac x2cosfrac x2=1-cos x=2sin^2frac x2$$



                  $$implies sinfrac x2left(cosfrac x2-sinfrac x2right)=0$$



                  $$(i)sinfrac x2=0implies frac x2=npitext{ where }n text{ is any integer}$$



                  $$(ii) cosfrac x2-sinfrac x2=0impliescosfrac x2=sinfrac x2iff tanfrac x2=1$$ Find the rest in my other answer






                  share|cite|improve this answer












                  Using Double-Angle Formulas,



                  $$sin x+cos x=1implies 2sinfrac x2cosfrac x2=1-cos x=2sin^2frac x2$$



                  $$implies sinfrac x2left(cosfrac x2-sinfrac x2right)=0$$



                  $$(i)sinfrac x2=0implies frac x2=npitext{ where }n text{ is any integer}$$



                  $$(ii) cosfrac x2-sinfrac x2=0impliescosfrac x2=sinfrac x2iff tanfrac x2=1$$ Find the rest in my other answer







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 26 '13 at 13:50









                  lab bhattacharjee

                  223k15156274




                  223k15156274























                      1














                      $sin x + cos x = 1 rightarrow (sin x + cos x )^2=1^2 iff color{green}{underbrace{(sin^2 x + cos^2 x)}_{=1}}+2sin x cos x =color{green}{1} iff sin x cos x = 0 $



                      Now look at the unit circle to see when this is true. After you have found the values of $x$ for which this holds, be sure to check the outcome of $sin x + cos x$ for each of these values, to eliminate the outcomes that generate value $-1$ (these values of $x$ came in because of the squaring - but are not truly solutions).






                      share|cite|improve this answer























                      • @DanielFischer My bad. Good point.
                        – dreamer
                        Dec 25 '13 at 18:27










                      • Note that squaring may add ghost solutions. You need to mention this. If we start with $x=1$ then there is one solution: $x=1$. Squaring both sides gives $x^2=1$, which has two solutions: $x=pm 1$. One solution, $x=1$, is a correct solution of the original problem while the other, $x=-1$, is a ghost solution introduced by squaring.
                        – Fly by Night
                        Dec 25 '13 at 19:17












                      • @FlybyNight You're right. Therefore, it would be good indeed to check the answer.
                        – dreamer
                        Dec 25 '13 at 19:22










                      • The equation $cos x sin x =0$ has $x=180^{circ}$ and $x=270^{circ}$ as solutions. However $sin(180^{circ})+cos(180^{circ})=-1$ and $sin(270^{circ})+cos(270^{circ})=-1$. The OP asked for solutions to $sin x + cos x =1$.
                        – Fly by Night
                        Dec 25 '13 at 19:46










                      • @FlybyNight No, those are not the only ones. $x=90^{circ}$ and $x=0^{circ}$ are also solutions , which both result in $sin x + cos x =1$.
                        – dreamer
                        Dec 25 '13 at 20:10
















                      1














                      $sin x + cos x = 1 rightarrow (sin x + cos x )^2=1^2 iff color{green}{underbrace{(sin^2 x + cos^2 x)}_{=1}}+2sin x cos x =color{green}{1} iff sin x cos x = 0 $



                      Now look at the unit circle to see when this is true. After you have found the values of $x$ for which this holds, be sure to check the outcome of $sin x + cos x$ for each of these values, to eliminate the outcomes that generate value $-1$ (these values of $x$ came in because of the squaring - but are not truly solutions).






                      share|cite|improve this answer























                      • @DanielFischer My bad. Good point.
                        – dreamer
                        Dec 25 '13 at 18:27










                      • Note that squaring may add ghost solutions. You need to mention this. If we start with $x=1$ then there is one solution: $x=1$. Squaring both sides gives $x^2=1$, which has two solutions: $x=pm 1$. One solution, $x=1$, is a correct solution of the original problem while the other, $x=-1$, is a ghost solution introduced by squaring.
                        – Fly by Night
                        Dec 25 '13 at 19:17












                      • @FlybyNight You're right. Therefore, it would be good indeed to check the answer.
                        – dreamer
                        Dec 25 '13 at 19:22










                      • The equation $cos x sin x =0$ has $x=180^{circ}$ and $x=270^{circ}$ as solutions. However $sin(180^{circ})+cos(180^{circ})=-1$ and $sin(270^{circ})+cos(270^{circ})=-1$. The OP asked for solutions to $sin x + cos x =1$.
                        – Fly by Night
                        Dec 25 '13 at 19:46










                      • @FlybyNight No, those are not the only ones. $x=90^{circ}$ and $x=0^{circ}$ are also solutions , which both result in $sin x + cos x =1$.
                        – dreamer
                        Dec 25 '13 at 20:10














                      1












                      1








                      1






                      $sin x + cos x = 1 rightarrow (sin x + cos x )^2=1^2 iff color{green}{underbrace{(sin^2 x + cos^2 x)}_{=1}}+2sin x cos x =color{green}{1} iff sin x cos x = 0 $



                      Now look at the unit circle to see when this is true. After you have found the values of $x$ for which this holds, be sure to check the outcome of $sin x + cos x$ for each of these values, to eliminate the outcomes that generate value $-1$ (these values of $x$ came in because of the squaring - but are not truly solutions).






                      share|cite|improve this answer














                      $sin x + cos x = 1 rightarrow (sin x + cos x )^2=1^2 iff color{green}{underbrace{(sin^2 x + cos^2 x)}_{=1}}+2sin x cos x =color{green}{1} iff sin x cos x = 0 $



                      Now look at the unit circle to see when this is true. After you have found the values of $x$ for which this holds, be sure to check the outcome of $sin x + cos x$ for each of these values, to eliminate the outcomes that generate value $-1$ (these values of $x$ came in because of the squaring - but are not truly solutions).







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 25 '13 at 20:49

























                      answered Dec 25 '13 at 18:25









                      dreamer

                      1,66643060




                      1,66643060












                      • @DanielFischer My bad. Good point.
                        – dreamer
                        Dec 25 '13 at 18:27










                      • Note that squaring may add ghost solutions. You need to mention this. If we start with $x=1$ then there is one solution: $x=1$. Squaring both sides gives $x^2=1$, which has two solutions: $x=pm 1$. One solution, $x=1$, is a correct solution of the original problem while the other, $x=-1$, is a ghost solution introduced by squaring.
                        – Fly by Night
                        Dec 25 '13 at 19:17












                      • @FlybyNight You're right. Therefore, it would be good indeed to check the answer.
                        – dreamer
                        Dec 25 '13 at 19:22










                      • The equation $cos x sin x =0$ has $x=180^{circ}$ and $x=270^{circ}$ as solutions. However $sin(180^{circ})+cos(180^{circ})=-1$ and $sin(270^{circ})+cos(270^{circ})=-1$. The OP asked for solutions to $sin x + cos x =1$.
                        – Fly by Night
                        Dec 25 '13 at 19:46










                      • @FlybyNight No, those are not the only ones. $x=90^{circ}$ and $x=0^{circ}$ are also solutions , which both result in $sin x + cos x =1$.
                        – dreamer
                        Dec 25 '13 at 20:10


















                      • @DanielFischer My bad. Good point.
                        – dreamer
                        Dec 25 '13 at 18:27










                      • Note that squaring may add ghost solutions. You need to mention this. If we start with $x=1$ then there is one solution: $x=1$. Squaring both sides gives $x^2=1$, which has two solutions: $x=pm 1$. One solution, $x=1$, is a correct solution of the original problem while the other, $x=-1$, is a ghost solution introduced by squaring.
                        – Fly by Night
                        Dec 25 '13 at 19:17












                      • @FlybyNight You're right. Therefore, it would be good indeed to check the answer.
                        – dreamer
                        Dec 25 '13 at 19:22










                      • The equation $cos x sin x =0$ has $x=180^{circ}$ and $x=270^{circ}$ as solutions. However $sin(180^{circ})+cos(180^{circ})=-1$ and $sin(270^{circ})+cos(270^{circ})=-1$. The OP asked for solutions to $sin x + cos x =1$.
                        – Fly by Night
                        Dec 25 '13 at 19:46










                      • @FlybyNight No, those are not the only ones. $x=90^{circ}$ and $x=0^{circ}$ are also solutions , which both result in $sin x + cos x =1$.
                        – dreamer
                        Dec 25 '13 at 20:10
















                      @DanielFischer My bad. Good point.
                      – dreamer
                      Dec 25 '13 at 18:27




                      @DanielFischer My bad. Good point.
                      – dreamer
                      Dec 25 '13 at 18:27












                      Note that squaring may add ghost solutions. You need to mention this. If we start with $x=1$ then there is one solution: $x=1$. Squaring both sides gives $x^2=1$, which has two solutions: $x=pm 1$. One solution, $x=1$, is a correct solution of the original problem while the other, $x=-1$, is a ghost solution introduced by squaring.
                      – Fly by Night
                      Dec 25 '13 at 19:17






                      Note that squaring may add ghost solutions. You need to mention this. If we start with $x=1$ then there is one solution: $x=1$. Squaring both sides gives $x^2=1$, which has two solutions: $x=pm 1$. One solution, $x=1$, is a correct solution of the original problem while the other, $x=-1$, is a ghost solution introduced by squaring.
                      – Fly by Night
                      Dec 25 '13 at 19:17














                      @FlybyNight You're right. Therefore, it would be good indeed to check the answer.
                      – dreamer
                      Dec 25 '13 at 19:22




                      @FlybyNight You're right. Therefore, it would be good indeed to check the answer.
                      – dreamer
                      Dec 25 '13 at 19:22












                      The equation $cos x sin x =0$ has $x=180^{circ}$ and $x=270^{circ}$ as solutions. However $sin(180^{circ})+cos(180^{circ})=-1$ and $sin(270^{circ})+cos(270^{circ})=-1$. The OP asked for solutions to $sin x + cos x =1$.
                      – Fly by Night
                      Dec 25 '13 at 19:46




                      The equation $cos x sin x =0$ has $x=180^{circ}$ and $x=270^{circ}$ as solutions. However $sin(180^{circ})+cos(180^{circ})=-1$ and $sin(270^{circ})+cos(270^{circ})=-1$. The OP asked for solutions to $sin x + cos x =1$.
                      – Fly by Night
                      Dec 25 '13 at 19:46












                      @FlybyNight No, those are not the only ones. $x=90^{circ}$ and $x=0^{circ}$ are also solutions , which both result in $sin x + cos x =1$.
                      – dreamer
                      Dec 25 '13 at 20:10




                      @FlybyNight No, those are not the only ones. $x=90^{circ}$ and $x=0^{circ}$ are also solutions , which both result in $sin x + cos x =1$.
                      – dreamer
                      Dec 25 '13 at 20:10


















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