Let $3sin x +cos x =2 $ then $frac{3sin x}{4sin x+3cos x}=?$
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
add a comment |
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
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
add a comment |
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
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
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
add a comment |
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
add a comment |
3 Answers
3
active
oldest
votes
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
$${3sin xover4sin x+3cos x}={3sin xover4sin x+3(2-3sin x)}={3sin xover6-5sin x}={3(6pmsqrt6)over60-5(6pmsqrt6)}={3(6pmsqrt6)over5(6mpsqrt6)}={3(6pmsqrt6)^2over5cdot30}\={21pm6sqrt6over25}$$
Other nice way too.
– gimusi
Nov 30 '18 at 22:03
add a comment |
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 $$t=1pm sqrt{frac 23}$$
@gimusi Ok . i did it. thanks .
– hamam_Abdallah
Nov 30 '18 at 22:04
add a comment |
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
$$3sin x +cos x =2iff cos xleft(3tan x+1right)=2 iff 3tan x+1=pm 2sqrt{1+tan^2x}$$
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
add a comment |
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3 Answers
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3 Answers
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votes
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
$${3sin xover4sin x+3cos x}={3sin xover4sin x+3(2-3sin x)}={3sin xover6-5sin x}={3(6pmsqrt6)over60-5(6pmsqrt6)}={3(6pmsqrt6)over5(6mpsqrt6)}={3(6pmsqrt6)^2over5cdot30}\={21pm6sqrt6over25}$$
Other nice way too.
– gimusi
Nov 30 '18 at 22:03
add a comment |
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
$${3sin xover4sin x+3cos x}={3sin xover4sin x+3(2-3sin x)}={3sin xover6-5sin x}={3(6pmsqrt6)over60-5(6pmsqrt6)}={3(6pmsqrt6)over5(6mpsqrt6)}={3(6pmsqrt6)^2over5cdot30}\={21pm6sqrt6over25}$$
Other nice way too.
– gimusi
Nov 30 '18 at 22:03
add a comment |
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
$${3sin xover4sin x+3cos x}={3sin xover4sin x+3(2-3sin x)}={3sin xover6-5sin x}={3(6pmsqrt6)over60-5(6pmsqrt6)}={3(6pmsqrt6)over5(6mpsqrt6)}={3(6pmsqrt6)^2over5cdot30}\={21pm6sqrt6over25}$$
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
$${3sin xover4sin x+3cos x}={3sin xover4sin x+3(2-3sin x)}={3sin xover6-5sin x}={3(6pmsqrt6)over60-5(6pmsqrt6)}={3(6pmsqrt6)over5(6mpsqrt6)}={3(6pmsqrt6)^2over5cdot30}\={21pm6sqrt6over25}$$
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
add a comment |
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
add a comment |
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 $$t=1pm sqrt{frac 23}$$
@gimusi Ok . i did it. thanks .
– hamam_Abdallah
Nov 30 '18 at 22:04
add a comment |
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 $$t=1pm sqrt{frac 23}$$
@gimusi Ok . i did it. thanks .
– hamam_Abdallah
Nov 30 '18 at 22:04
add a comment |
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 $$t=1pm sqrt{frac 23}$$
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 $$t=1pm sqrt{frac 23}$$
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
add a comment |
@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
add a comment |
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
$$3sin x +cos x =2iff cos xleft(3tan x+1right)=2 iff 3tan x+1=pm 2sqrt{1+tan^2x}$$
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
add a comment |
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
$$3sin x +cos x =2iff cos xleft(3tan x+1right)=2 iff 3tan x+1=pm 2sqrt{1+tan^2x}$$
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
add a comment |
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
$$3sin x +cos x =2iff cos xleft(3tan x+1right)=2 iff 3tan x+1=pm 2sqrt{1+tan^2x}$$
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
$$3sin x +cos x =2iff cos xleft(3tan x+1right)=2 iff 3tan x+1=pm 2sqrt{1+tan^2x}$$
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
add a comment |
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
add a comment |
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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