Let $3sin x +cos x =2 $ then $frac{3sin x}{4sin x+3cos x}=?$












3














Let $3sin x +cos x =2 $ then $dfrac{3sin x}{4sin x+3cos x}=,?$




My try :



$$frac{frac{3sin x}{cos x}}{frac{4sin x+3cos x}{cos x}}=frac{3tan x}{4tan x+3} =;?$$



Now we have to find $tan x$ from $3sin x +cos x =2 $ but how?






calculus







share|cite|improve this question
























  • Let $sin(x)=s$ and $cos(x)=sqrt{1-s^2}$. Then solve for $s$. Alternatively, note $3sin(x)+cos(x)=sqrt{10}cos(x-arctan(sqrt{10}/5)$, solve for $x$ and substitute.
    – Mark Viola
    Nov 30 '18 at 20:43
















3














Let $3sin x +cos x =2 $ then $dfrac{3sin x}{4sin x+3cos x}=,?$




My try :



$$frac{frac{3sin x}{cos x}}{frac{4sin x+3cos x}{cos x}}=frac{3tan x}{4tan x+3} =;?$$



Now we have to find $tan x$ from $3sin x +cos x =2 $ but how?






calculus







share|cite|improve this question
























  • Let $sin(x)=s$ and $cos(x)=sqrt{1-s^2}$. Then solve for $s$. Alternatively, note $3sin(x)+cos(x)=sqrt{10}cos(x-arctan(sqrt{10}/5)$, solve for $x$ and substitute.
    – Mark Viola
    Nov 30 '18 at 20:43














3












3








3







Let $3sin x +cos x =2 $ then $dfrac{3sin x}{4sin x+3cos x}=,?$




My try :



$$frac{frac{3sin x}{cos x}}{frac{4sin x+3cos x}{cos x}}=frac{3tan x}{4tan x+3} =;?$$



Now we have to find $tan x$ from $3sin x +cos x =2 $ but how?






calculus







share|cite|improve this question















Let $3sin x +cos x =2 $ then $dfrac{3sin x}{4sin x+3cos x}=,?$




My try :



$$frac{frac{3sin x}{cos x}}{frac{4sin x+3cos x}{cos x}}=frac{3tan x}{4tan x+3} =;?$$



Now we have to find $tan x$ from $3sin x +cos x =2 $ but how?






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 20:02









amWhy

192k28225439




192k28225439










asked Nov 30 '18 at 19:22









Almot1960Almot1960

2,511823




2,511823












  • Let $sin(x)=s$ and $cos(x)=sqrt{1-s^2}$. Then solve for $s$. Alternatively, note $3sin(x)+cos(x)=sqrt{10}cos(x-arctan(sqrt{10}/5)$, solve for $x$ and substitute.
    – Mark Viola
    Nov 30 '18 at 20:43


















  • Let $sin(x)=s$ and $cos(x)=sqrt{1-s^2}$. Then solve for $s$. Alternatively, note $3sin(x)+cos(x)=sqrt{10}cos(x-arctan(sqrt{10}/5)$, solve for $x$ and substitute.
    – Mark Viola
    Nov 30 '18 at 20:43
















Let $sin(x)=s$ and $cos(x)=sqrt{1-s^2}$. Then solve for $s$. Alternatively, note $3sin(x)+cos(x)=sqrt{10}cos(x-arctan(sqrt{10}/5)$, solve for $x$ and substitute.
– Mark Viola
Nov 30 '18 at 20:43




Let $sin(x)=s$ and $cos(x)=sqrt{1-s^2}$. Then solve for $s$. Alternatively, note $3sin(x)+cos(x)=sqrt{10}cos(x-arctan(sqrt{10}/5)$, solve for $x$ and substitute.
– Mark Viola
Nov 30 '18 at 20:43










3 Answers
3






active

oldest

votes


















2














Instead of working with tangents, I'd recommend getting rid of the cosine and solving for $sin x$:



$3sin x+cos x=2$ implies $1-sin^2x=cos^2x=(2-3sin x)^2=4-12sin x+9sin^2x$, or



$$10sin^2x-12sin x+3=0$$



which solves to $sin x=(6pmsqrt{36-30})/10=(6pmsqrt6)/10$. Both are valid solutions, since $|sin x|=|6pmsqrt6|/10$ and $|cos x|=|2-3sin x|=|2mp3sqrt6|/10$ are less than or equal to $1$ for both signs.



We now have



$${3sin xover4sin x+3cos x}={3sin xover4sin x+3(2-3sin x)}={3sin xover6-5sin x}={3(6pmsqrt6)over60-5(6pmsqrt6)}={3(6pmsqrt6)over5(6mpsqrt6)}={3(6pmsqrt6)^2over5cdot30}\={21pm6sqrt6over25}$$






share|cite|improve this answer





















  • Other nice way too.
    – gimusi
    Nov 30 '18 at 22:03



















2














Let $t=tan(frac x2)$.



use the identities



$$sin(x)=frac{2t}{1+t^2}$$



$$cos(x)=frac{1-t^2}{1+t^2},$$
and
$$tan(x)=frac{2t}{1-t^2}.$$



thus
$$3sin(x)+cos(x)=2implies$$
$$6t+1-t^2=2(1+t^2) implies$$
or
$$3t^2-6t+1=0$$



hence
$$frac{3tan(x)}{4tan(x)+3}=$$



$$frac{6t}{8t+3(1-t^2)}=$$
$$frac{6t}{2t+4}=3-frac{6}{t+2}$$



with $$t=1pm sqrt{frac 23}$$






share|cite|improve this answer























  • @gimusi Ok . i did it. thanks .
    – hamam_Abdallah
    Nov 30 '18 at 22:04



















0














HINT



By tangent half-angle substitution with $t=tan frac x 2$ we have



$$3sin x +cos x =2 iff 3frac{2t}{1+t^2}+frac{1-t^2}{1+t^2}=2 iff 3t^2-6t+1=0$$



then



$$dfrac{3sin x}{4sin x+3cos x}=frac{6t}{-3t^2+8t+3}=frac{6t}{(-3t^2-1)+8t+4}=frac{6t}{2t+4}=frac{3t}{t+2}$$



As an alternative since $cos x=pm frac1{sqrt{1+tan^2x}}$ we can use that



$$3sin x +cos x =2iff cos xleft(3tan x+1right)=2 iff 3tan x+1=pm 2sqrt{1+tan^2x}$$






share|cite|improve this answer























  • Your last denominator is not correct.
    – hamam_Abdallah
    Nov 30 '18 at 19:43










  • @hamam_Abdallah Yes of course, I fix the typo. Thanks
    – gimusi
    Nov 30 '18 at 19:44











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














Instead of working with tangents, I'd recommend getting rid of the cosine and solving for $sin x$:



$3sin x+cos x=2$ implies $1-sin^2x=cos^2x=(2-3sin x)^2=4-12sin x+9sin^2x$, or



$$10sin^2x-12sin x+3=0$$



which solves to $sin x=(6pmsqrt{36-30})/10=(6pmsqrt6)/10$. Both are valid solutions, since $|sin x|=|6pmsqrt6|/10$ and $|cos x|=|2-3sin x|=|2mp3sqrt6|/10$ are less than or equal to $1$ for both signs.



We now have



$${3sin xover4sin x+3cos x}={3sin xover4sin x+3(2-3sin x)}={3sin xover6-5sin x}={3(6pmsqrt6)over60-5(6pmsqrt6)}={3(6pmsqrt6)over5(6mpsqrt6)}={3(6pmsqrt6)^2over5cdot30}\={21pm6sqrt6over25}$$






share|cite|improve this answer





















  • Other nice way too.
    – gimusi
    Nov 30 '18 at 22:03
















2














Instead of working with tangents, I'd recommend getting rid of the cosine and solving for $sin x$:



$3sin x+cos x=2$ implies $1-sin^2x=cos^2x=(2-3sin x)^2=4-12sin x+9sin^2x$, or



$$10sin^2x-12sin x+3=0$$



which solves to $sin x=(6pmsqrt{36-30})/10=(6pmsqrt6)/10$. Both are valid solutions, since $|sin x|=|6pmsqrt6|/10$ and $|cos x|=|2-3sin x|=|2mp3sqrt6|/10$ are less than or equal to $1$ for both signs.



We now have



$${3sin xover4sin x+3cos x}={3sin xover4sin x+3(2-3sin x)}={3sin xover6-5sin x}={3(6pmsqrt6)over60-5(6pmsqrt6)}={3(6pmsqrt6)over5(6mpsqrt6)}={3(6pmsqrt6)^2over5cdot30}\={21pm6sqrt6over25}$$






share|cite|improve this answer





















  • Other nice way too.
    – gimusi
    Nov 30 '18 at 22:03














2












2








2






Instead of working with tangents, I'd recommend getting rid of the cosine and solving for $sin x$:



$3sin x+cos x=2$ implies $1-sin^2x=cos^2x=(2-3sin x)^2=4-12sin x+9sin^2x$, or



$$10sin^2x-12sin x+3=0$$



which solves to $sin x=(6pmsqrt{36-30})/10=(6pmsqrt6)/10$. Both are valid solutions, since $|sin x|=|6pmsqrt6|/10$ and $|cos x|=|2-3sin x|=|2mp3sqrt6|/10$ are less than or equal to $1$ for both signs.



We now have



$${3sin xover4sin x+3cos x}={3sin xover4sin x+3(2-3sin x)}={3sin xover6-5sin x}={3(6pmsqrt6)over60-5(6pmsqrt6)}={3(6pmsqrt6)over5(6mpsqrt6)}={3(6pmsqrt6)^2over5cdot30}\={21pm6sqrt6over25}$$






share|cite|improve this answer












Instead of working with tangents, I'd recommend getting rid of the cosine and solving for $sin x$:



$3sin x+cos x=2$ implies $1-sin^2x=cos^2x=(2-3sin x)^2=4-12sin x+9sin^2x$, or



$$10sin^2x-12sin x+3=0$$



which solves to $sin x=(6pmsqrt{36-30})/10=(6pmsqrt6)/10$. Both are valid solutions, since $|sin x|=|6pmsqrt6|/10$ and $|cos x|=|2-3sin x|=|2mp3sqrt6|/10$ are less than or equal to $1$ for both signs.



We now have



$${3sin xover4sin x+3cos x}={3sin xover4sin x+3(2-3sin x)}={3sin xover6-5sin x}={3(6pmsqrt6)over60-5(6pmsqrt6)}={3(6pmsqrt6)over5(6mpsqrt6)}={3(6pmsqrt6)^2over5cdot30}\={21pm6sqrt6over25}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 30 '18 at 21:34









Barry CipraBarry Cipra

59.3k653125




59.3k653125












  • Other nice way too.
    – gimusi
    Nov 30 '18 at 22:03


















  • Other nice way too.
    – gimusi
    Nov 30 '18 at 22:03
















Other nice way too.
– gimusi
Nov 30 '18 at 22:03




Other nice way too.
– gimusi
Nov 30 '18 at 22:03











2














Let $t=tan(frac x2)$.



use the identities



$$sin(x)=frac{2t}{1+t^2}$$



$$cos(x)=frac{1-t^2}{1+t^2},$$
and
$$tan(x)=frac{2t}{1-t^2}.$$



thus
$$3sin(x)+cos(x)=2implies$$
$$6t+1-t^2=2(1+t^2) implies$$
or
$$3t^2-6t+1=0$$



hence
$$frac{3tan(x)}{4tan(x)+3}=$$



$$frac{6t}{8t+3(1-t^2)}=$$
$$frac{6t}{2t+4}=3-frac{6}{t+2}$$



with $$t=1pm sqrt{frac 23}$$






share|cite|improve this answer























  • @gimusi Ok . i did it. thanks .
    – hamam_Abdallah
    Nov 30 '18 at 22:04
















2














Let $t=tan(frac x2)$.



use the identities



$$sin(x)=frac{2t}{1+t^2}$$



$$cos(x)=frac{1-t^2}{1+t^2},$$
and
$$tan(x)=frac{2t}{1-t^2}.$$



thus
$$3sin(x)+cos(x)=2implies$$
$$6t+1-t^2=2(1+t^2) implies$$
or
$$3t^2-6t+1=0$$



hence
$$frac{3tan(x)}{4tan(x)+3}=$$



$$frac{6t}{8t+3(1-t^2)}=$$
$$frac{6t}{2t+4}=3-frac{6}{t+2}$$



with $$t=1pm sqrt{frac 23}$$






share|cite|improve this answer























  • @gimusi Ok . i did it. thanks .
    – hamam_Abdallah
    Nov 30 '18 at 22:04














2












2








2






Let $t=tan(frac x2)$.



use the identities



$$sin(x)=frac{2t}{1+t^2}$$



$$cos(x)=frac{1-t^2}{1+t^2},$$
and
$$tan(x)=frac{2t}{1-t^2}.$$



thus
$$3sin(x)+cos(x)=2implies$$
$$6t+1-t^2=2(1+t^2) implies$$
or
$$3t^2-6t+1=0$$



hence
$$frac{3tan(x)}{4tan(x)+3}=$$



$$frac{6t}{8t+3(1-t^2)}=$$
$$frac{6t}{2t+4}=3-frac{6}{t+2}$$



with $$t=1pm sqrt{frac 23}$$






share|cite|improve this answer














Let $t=tan(frac x2)$.



use the identities



$$sin(x)=frac{2t}{1+t^2}$$



$$cos(x)=frac{1-t^2}{1+t^2},$$
and
$$tan(x)=frac{2t}{1-t^2}.$$



thus
$$3sin(x)+cos(x)=2implies$$
$$6t+1-t^2=2(1+t^2) implies$$
or
$$3t^2-6t+1=0$$



hence
$$frac{3tan(x)}{4tan(x)+3}=$$



$$frac{6t}{8t+3(1-t^2)}=$$
$$frac{6t}{2t+4}=3-frac{6}{t+2}$$



with $$t=1pm sqrt{frac 23}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 30 '18 at 22:03

























answered Nov 30 '18 at 19:34









hamam_Abdallahhamam_Abdallah

38k21634




38k21634












  • @gimusi Ok . i did it. thanks .
    – hamam_Abdallah
    Nov 30 '18 at 22:04


















  • @gimusi Ok . i did it. thanks .
    – hamam_Abdallah
    Nov 30 '18 at 22:04
















@gimusi Ok . i did it. thanks .
– hamam_Abdallah
Nov 30 '18 at 22:04




@gimusi Ok . i did it. thanks .
– hamam_Abdallah
Nov 30 '18 at 22:04











0














HINT



By tangent half-angle substitution with $t=tan frac x 2$ we have



$$3sin x +cos x =2 iff 3frac{2t}{1+t^2}+frac{1-t^2}{1+t^2}=2 iff 3t^2-6t+1=0$$



then



$$dfrac{3sin x}{4sin x+3cos x}=frac{6t}{-3t^2+8t+3}=frac{6t}{(-3t^2-1)+8t+4}=frac{6t}{2t+4}=frac{3t}{t+2}$$



As an alternative since $cos x=pm frac1{sqrt{1+tan^2x}}$ we can use that



$$3sin x +cos x =2iff cos xleft(3tan x+1right)=2 iff 3tan x+1=pm 2sqrt{1+tan^2x}$$






share|cite|improve this answer























  • Your last denominator is not correct.
    – hamam_Abdallah
    Nov 30 '18 at 19:43










  • @hamam_Abdallah Yes of course, I fix the typo. Thanks
    – gimusi
    Nov 30 '18 at 19:44
















0














HINT



By tangent half-angle substitution with $t=tan frac x 2$ we have



$$3sin x +cos x =2 iff 3frac{2t}{1+t^2}+frac{1-t^2}{1+t^2}=2 iff 3t^2-6t+1=0$$



then



$$dfrac{3sin x}{4sin x+3cos x}=frac{6t}{-3t^2+8t+3}=frac{6t}{(-3t^2-1)+8t+4}=frac{6t}{2t+4}=frac{3t}{t+2}$$



As an alternative since $cos x=pm frac1{sqrt{1+tan^2x}}$ we can use that



$$3sin x +cos x =2iff cos xleft(3tan x+1right)=2 iff 3tan x+1=pm 2sqrt{1+tan^2x}$$






share|cite|improve this answer























  • Your last denominator is not correct.
    – hamam_Abdallah
    Nov 30 '18 at 19:43










  • @hamam_Abdallah Yes of course, I fix the typo. Thanks
    – gimusi
    Nov 30 '18 at 19:44














0












0








0






HINT



By tangent half-angle substitution with $t=tan frac x 2$ we have



$$3sin x +cos x =2 iff 3frac{2t}{1+t^2}+frac{1-t^2}{1+t^2}=2 iff 3t^2-6t+1=0$$



then



$$dfrac{3sin x}{4sin x+3cos x}=frac{6t}{-3t^2+8t+3}=frac{6t}{(-3t^2-1)+8t+4}=frac{6t}{2t+4}=frac{3t}{t+2}$$



As an alternative since $cos x=pm frac1{sqrt{1+tan^2x}}$ we can use that



$$3sin x +cos x =2iff cos xleft(3tan x+1right)=2 iff 3tan x+1=pm 2sqrt{1+tan^2x}$$






share|cite|improve this answer














HINT



By tangent half-angle substitution with $t=tan frac x 2$ we have



$$3sin x +cos x =2 iff 3frac{2t}{1+t^2}+frac{1-t^2}{1+t^2}=2 iff 3t^2-6t+1=0$$



then



$$dfrac{3sin x}{4sin x+3cos x}=frac{6t}{-3t^2+8t+3}=frac{6t}{(-3t^2-1)+8t+4}=frac{6t}{2t+4}=frac{3t}{t+2}$$



As an alternative since $cos x=pm frac1{sqrt{1+tan^2x}}$ we can use that



$$3sin x +cos x =2iff cos xleft(3tan x+1right)=2 iff 3tan x+1=pm 2sqrt{1+tan^2x}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 30 '18 at 21:41

























answered Nov 30 '18 at 19:32









gimusigimusi

1




1












  • Your last denominator is not correct.
    – hamam_Abdallah
    Nov 30 '18 at 19:43










  • @hamam_Abdallah Yes of course, I fix the typo. Thanks
    – gimusi
    Nov 30 '18 at 19:44


















  • Your last denominator is not correct.
    – hamam_Abdallah
    Nov 30 '18 at 19:43










  • @hamam_Abdallah Yes of course, I fix the typo. Thanks
    – gimusi
    Nov 30 '18 at 19:44
















Your last denominator is not correct.
– hamam_Abdallah
Nov 30 '18 at 19:43




Your last denominator is not correct.
– hamam_Abdallah
Nov 30 '18 at 19:43












@hamam_Abdallah Yes of course, I fix the typo. Thanks
– gimusi
Nov 30 '18 at 19:44




@hamam_Abdallah Yes of course, I fix the typo. Thanks
– gimusi
Nov 30 '18 at 19:44


















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