How to find the roots of $fracsqrt3-1sin x+fracsqrt3+1cos x=4sqrt2$?How to integrate $sqrt1-sin 2x$?Show that $5cos^2x - 2sqrt3sin xcos x + 3sin^2x = cos 2x - sqrt3sin2x + 4$How to Simplify $-frac14sinfrac x4 + fracsqrt34cosfrac x4 = 0$?Number of solutions of $8sin(x)=fracsqrt3cos(x)+frac1sin(x)$Solve $2cos^2x=sqrt3sin2x$.If $fracsin xsin y= 3$ and $fraccos xcos y= frac12$, then find $fracsin2xsin2y+fraccos2xcos2y$Find the general value of $x$ satisfying $sin x=frac12$ and $cos x=-fracsqrt 32$How to solve $sin^2(x)+sin2x+2cos^2(x)=0$Area bounded by $y=sqrtfrac1+sinxcosx$ and $y=sqrtfrac1-sinxcosx$Interval in which $(cos p-1)x^2+cos p.x+sin p=0$ has real roots
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How to find the roots of $fracsqrt3-1sin x+fracsqrt3+1cos x=4sqrt2$?
How to integrate $sqrt1-sin 2x$?Show that $5cos^2x - 2sqrt3sin xcos x + 3sin^2x = cos 2x - sqrt3sin2x + 4$How to Simplify $-frac14sinfrac x4 + fracsqrt34cosfrac x4 = 0$?Number of solutions of $8sin(x)=fracsqrt3cos(x)+frac1sin(x)$Solve $2cos^2x=sqrt3sin2x$.If $fracsin xsin y= 3$ and $fraccos xcos y= frac12$, then find $fracsin2xsin2y+fraccos2xcos2y$Find the general value of $x$ satisfying $sin x=frac12$ and $cos x=-fracsqrt 32$How to solve $sin^2(x)+sin2x+2cos^2(x)=0$Area bounded by $y=sqrtfrac1+sinxcosx$ and $y=sqrtfrac1-sinxcosx$Interval in which $(cos p-1)x^2+cos p.x+sin p=0$ has real roots
$begingroup$
Find all $x$ in the interval $(0,pi/2)$ such that $fracsqrt3-1sin x+fracsqrt3+1cos x=4sqrt2$.
The options are (i)$pi/9,2pi/7$, (ii)$pi/36,11pi/12$ (iii)$pi/12,11pi/36$ (iv) All
I have been able to find one value of $x$, $pi/12$. How do I find the other root(s)?
My attempt:
$fracsqrt3-1sin x+fracsqrt3+1cos x=4sqrt2$
or, $fracsinpi/3-sinpi/6sin x+fraccospi/6+cospi/3cos x=2sqrt2$
or, $fracsin(pi/4)cos(pi/12)sin x+fraccos(pi/4)cos(pi/12)cos x=sqrt2$
or, $sin(x+pi/12)=sin2x$
or, $x=pi/12$
trigonometry
asked 2 days ago
MrAPMrAP
1,16721432
$endgroup$
add a comment |
$begingroup$
Find all $x$ in the interval $(0,pi/2)$ such that $fracsqrt3-1sin x+fracsqrt3+1cos x=4sqrt2$.
The options are (i)$pi/9,2pi/7$, (ii)$pi/36,11pi/12$ (iii)$pi/12,11pi/36$ (iv) All
I have been able to find one value of $x$, $pi/12$. How do I find the other root(s)?
My attempt:
$fracsqrt3-1sin x+fracsqrt3+1cos x=4sqrt2$
or, $fracsinpi/3-sinpi/6sin x+fraccospi/6+cospi/3cos x=2sqrt2$
or, $fracsin(pi/4)cos(pi/12)sin x+fraccos(pi/4)cos(pi/12)cos x=sqrt2$
or, $sin(x+pi/12)=sin2x$
or, $x=pi/12$
trigonometry
asked 2 days ago
MrAPMrAP
1,16721432
$endgroup$
1
$begingroup$
How did you find the one value that you found? (Perhaps show your working -- it's possible that you accidentally lost some solutions.)
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
I expressed the first numerator in terms of sine and the second in terms of cosine function. Then I used addition formula to combine both sines and cosines. After simplifying, I got $x=pi/12$
$endgroup$
– MrAP
2 days ago
1
$begingroup$
Maybe you accidentally missed some solutions when you got to the simplified stage. Remember that if $sin a = sin b$, then all the solutions (plural!) or $a$ in terms of $b$ are $a = b + 2npi$ or $a = pi - b + 2npi$, for some integer $n$. You will want to take only those solutions for which $x$ comes out to be between $0$ and $pi/2$.
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
@MinusOne-Twelfth, Check my question now.
$endgroup$
– MrAP
2 days ago
1
$begingroup$
You have simply equated the angles. As I said in my previous comment, there are more solutions to $sin a = sin b$ than just $a=b$. This is why you were unable to obtain all the solutions.
$endgroup$
– Minus One-Twelfth
2 days ago
add a comment |
$begingroup$
Find all $x$ in the interval $(0,pi/2)$ such that $fracsqrt3-1sin x+fracsqrt3+1cos x=4sqrt2$.
The options are (i)$pi/9,2pi/7$, (ii)$pi/36,11pi/12$ (iii)$pi/12,11pi/36$ (iv) All
I have been able to find one value of $x$, $pi/12$. How do I find the other root(s)?
My attempt:
$fracsqrt3-1sin x+fracsqrt3+1cos x=4sqrt2$
or, $fracsinpi/3-sinpi/6sin x+fraccospi/6+cospi/3cos x=2sqrt2$
or, $fracsin(pi/4)cos(pi/12)sin x+fraccos(pi/4)cos(pi/12)cos x=sqrt2$
or, $sin(x+pi/12)=sin2x$
or, $x=pi/12$
trigonometry
asked 2 days ago
MrAPMrAP
1,16721432
$endgroup$
Find all $x$ in the interval $(0,pi/2)$ such that $fracsqrt3-1sin x+fracsqrt3+1cos x=4sqrt2$.
The options are (i)$pi/9,2pi/7$, (ii)$pi/36,11pi/12$ (iii)$pi/12,11pi/36$ (iv) All
I have been able to find one value of $x$, $pi/12$. How do I find the other root(s)?
My attempt:
$fracsqrt3-1sin x+fracsqrt3+1cos x=4sqrt2$
or, $fracsinpi/3-sinpi/6sin x+fraccospi/6+cospi/3cos x=2sqrt2$
or, $fracsin(pi/4)cos(pi/12)sin x+fraccos(pi/4)cos(pi/12)cos x=sqrt2$
or, $sin(x+pi/12)=sin2x$
or, $x=pi/12$
trigonometry
trigonometry
asked 2 days ago
MrAPMrAP
1,16721432
asked 2 days ago
MrAPMrAP
1,16721432
edited 2 days ago
MrAP
asked 2 days ago
MrAPMrAP
1,16721432
asked 2 days ago
MrAPMrAP
1,16721432
asked 2 days ago
MrAPMrAP
1,16721432
1,16721432
1
$begingroup$
How did you find the one value that you found? (Perhaps show your working -- it's possible that you accidentally lost some solutions.)
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
I expressed the first numerator in terms of sine and the second in terms of cosine function. Then I used addition formula to combine both sines and cosines. After simplifying, I got $x=pi/12$
$endgroup$
– MrAP
2 days ago
1
$begingroup$
Maybe you accidentally missed some solutions when you got to the simplified stage. Remember that if $sin a = sin b$, then all the solutions (plural!) or $a$ in terms of $b$ are $a = b + 2npi$ or $a = pi - b + 2npi$, for some integer $n$. You will want to take only those solutions for which $x$ comes out to be between $0$ and $pi/2$.
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
@MinusOne-Twelfth, Check my question now.
$endgroup$
– MrAP
2 days ago
1
$begingroup$
You have simply equated the angles. As I said in my previous comment, there are more solutions to $sin a = sin b$ than just $a=b$. This is why you were unable to obtain all the solutions.
$endgroup$
– Minus One-Twelfth
2 days ago
add a comment |
1
$begingroup$
How did you find the one value that you found? (Perhaps show your working -- it's possible that you accidentally lost some solutions.)
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
I expressed the first numerator in terms of sine and the second in terms of cosine function. Then I used addition formula to combine both sines and cosines. After simplifying, I got $x=pi/12$
$endgroup$
– MrAP
2 days ago
1
$begingroup$
Maybe you accidentally missed some solutions when you got to the simplified stage. Remember that if $sin a = sin b$, then all the solutions (plural!) or $a$ in terms of $b$ are $a = b + 2npi$ or $a = pi - b + 2npi$, for some integer $n$. You will want to take only those solutions for which $x$ comes out to be between $0$ and $pi/2$.
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
@MinusOne-Twelfth, Check my question now.
$endgroup$
– MrAP
2 days ago
1
$begingroup$
You have simply equated the angles. As I said in my previous comment, there are more solutions to $sin a = sin b$ than just $a=b$. This is why you were unable to obtain all the solutions.
$endgroup$
– Minus One-Twelfth
2 days ago
1
1
$begingroup$
How did you find the one value that you found? (Perhaps show your working -- it's possible that you accidentally lost some solutions.)
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
How did you find the one value that you found? (Perhaps show your working -- it's possible that you accidentally lost some solutions.)
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
I expressed the first numerator in terms of sine and the second in terms of cosine function. Then I used addition formula to combine both sines and cosines. After simplifying, I got $x=pi/12$
$endgroup$
– MrAP
2 days ago
$begingroup$
I expressed the first numerator in terms of sine and the second in terms of cosine function. Then I used addition formula to combine both sines and cosines. After simplifying, I got $x=pi/12$
$endgroup$
– MrAP
2 days ago
1
1
$begingroup$
Maybe you accidentally missed some solutions when you got to the simplified stage. Remember that if $sin a = sin b$, then all the solutions (plural!) or $a$ in terms of $b$ are $a = b + 2npi$ or $a = pi - b + 2npi$, for some integer $n$. You will want to take only those solutions for which $x$ comes out to be between $0$ and $pi/2$.
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
Maybe you accidentally missed some solutions when you got to the simplified stage. Remember that if $sin a = sin b$, then all the solutions (plural!) or $a$ in terms of $b$ are $a = b + 2npi$ or $a = pi - b + 2npi$, for some integer $n$. You will want to take only those solutions for which $x$ comes out to be between $0$ and $pi/2$.
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
@MinusOne-Twelfth, Check my question now.
$endgroup$
– MrAP
2 days ago
$begingroup$
@MinusOne-Twelfth, Check my question now.
$endgroup$
– MrAP
2 days ago
1
1
$begingroup$
You have simply equated the angles. As I said in my previous comment, there are more solutions to $sin a = sin b$ than just $a=b$. This is why you were unable to obtain all the solutions.
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
You have simply equated the angles. As I said in my previous comment, there are more solutions to $sin a = sin b$ than just $a=b$. This is why you were unable to obtain all the solutions.
$endgroup$
– Minus One-Twelfth
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Be careful that your final equation has more potential solutions. The equation
$$ sin left(x + fracpi12right) = sin 2x$$
implies in fact
$$ x + fracpi12 = 2x + 2k pi$$
or
$$ x + fracpi12 = pi - 2x + 2k pi.$$
Also recall that you can always check the number of solutions by intersecting
$$ fracsqrt 3 -1Y + fracsqrt 3 +1X=4 sqrt 2$$
with the unit circle
$$X^2+Y^2 = 1.$$
answered 2 days ago
MatteoMatteo
1,032313
$endgroup$
$begingroup$
How did you get the second and third equation from the first equation in your answer. I knew that the general solution of $sin x=sin a$ is $x=npi+(-1)^na$.
$endgroup$
– MrAP
2 days ago
$begingroup$
@MrAP. Consider that two supplementary angles have the same sine. So $$sin(pi- alpha) = sin alpha.$$
$endgroup$
– Matteo
2 days ago
1
$begingroup$
Oh. I got it. Mine generalizes both odd and even n in one equation.
$endgroup$
– MrAP
2 days ago
$begingroup$
@MrAP. Yep. I just prefer to use basic definitions instead of formulae. More ductile, I believe.
$endgroup$
– Matteo
2 days ago
add a comment |
$begingroup$
Use $$sin15^circ=fracsqrt3-12sqrt2$$ and
$$cos15^circ=fracsqrt3+12sqrt2.$$
We obtain:
$$sin(15^circ+x)=sin2x.$$
Thus, $$15^circ+x=2x+360^circk,$$ where $k$ is an integer number, or
$$15^circ+x=180^circ-2x+360^circk.$$
Can you end it now?
answered 2 days ago
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
I had got that equation at the end. Then I equated the angles to get $x=15^circ$
$endgroup$
– MrAP
2 days ago
$begingroup$
@MrAP I got also $55^circ.$
$endgroup$
– Michael Rozenberg
2 days ago
$begingroup$
How did you get that value?
$endgroup$
– MrAP
2 days ago
$begingroup$
@MrAP I added something. See now.
$endgroup$
– Michael Rozenberg
2 days ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Be careful that your final equation has more potential solutions. The equation
$$ sin left(x + fracpi12right) = sin 2x$$
implies in fact
$$ x + fracpi12 = 2x + 2k pi$$
or
$$ x + fracpi12 = pi - 2x + 2k pi.$$
Also recall that you can always check the number of solutions by intersecting
$$ fracsqrt 3 -1Y + fracsqrt 3 +1X=4 sqrt 2$$
with the unit circle
$$X^2+Y^2 = 1.$$
answered 2 days ago
MatteoMatteo
1,032313
$endgroup$
$begingroup$
How did you get the second and third equation from the first equation in your answer. I knew that the general solution of $sin x=sin a$ is $x=npi+(-1)^na$.
$endgroup$
– MrAP
2 days ago
$begingroup$
@MrAP. Consider that two supplementary angles have the same sine. So $$sin(pi- alpha) = sin alpha.$$
$endgroup$
– Matteo
2 days ago
1
$begingroup$
Oh. I got it. Mine generalizes both odd and even n in one equation.
$endgroup$
– MrAP
2 days ago
$begingroup$
@MrAP. Yep. I just prefer to use basic definitions instead of formulae. More ductile, I believe.
$endgroup$
– Matteo
2 days ago
add a comment |
$begingroup$
Be careful that your final equation has more potential solutions. The equation
$$ sin left(x + fracpi12right) = sin 2x$$
implies in fact
$$ x + fracpi12 = 2x + 2k pi$$
or
$$ x + fracpi12 = pi - 2x + 2k pi.$$
Also recall that you can always check the number of solutions by intersecting
$$ fracsqrt 3 -1Y + fracsqrt 3 +1X=4 sqrt 2$$
with the unit circle
$$X^2+Y^2 = 1.$$
answered 2 days ago
MatteoMatteo
1,032313
$endgroup$
$begingroup$
How did you get the second and third equation from the first equation in your answer. I knew that the general solution of $sin x=sin a$ is $x=npi+(-1)^na$.
$endgroup$
– MrAP
2 days ago
$begingroup$
@MrAP. Consider that two supplementary angles have the same sine. So $$sin(pi- alpha) = sin alpha.$$
$endgroup$
– Matteo
2 days ago
1
$begingroup$
Oh. I got it. Mine generalizes both odd and even n in one equation.
$endgroup$
– MrAP
2 days ago
$begingroup$
@MrAP. Yep. I just prefer to use basic definitions instead of formulae. More ductile, I believe.
$endgroup$
– Matteo
2 days ago
add a comment |
$begingroup$
Be careful that your final equation has more potential solutions. The equation
$$ sin left(x + fracpi12right) = sin 2x$$
implies in fact
$$ x + fracpi12 = 2x + 2k pi$$
or
$$ x + fracpi12 = pi - 2x + 2k pi.$$
Also recall that you can always check the number of solutions by intersecting
$$ fracsqrt 3 -1Y + fracsqrt 3 +1X=4 sqrt 2$$
with the unit circle
$$X^2+Y^2 = 1.$$
answered 2 days ago
MatteoMatteo
1,032313
$endgroup$
Be careful that your final equation has more potential solutions. The equation
$$ sin left(x + fracpi12right) = sin 2x$$
implies in fact
$$ x + fracpi12 = 2x + 2k pi$$
or
$$ x + fracpi12 = pi - 2x + 2k pi.$$
Also recall that you can always check the number of solutions by intersecting
$$ fracsqrt 3 -1Y + fracsqrt 3 +1X=4 sqrt 2$$
with the unit circle
$$X^2+Y^2 = 1.$$
answered 2 days ago
MatteoMatteo
1,032313
edited 2 days ago
answered 2 days ago
MatteoMatteo
1,032313
answered 2 days ago
MatteoMatteo
1,032313
answered 2 days ago
MatteoMatteo
1,032313
1,032313
$begingroup$
How did you get the second and third equation from the first equation in your answer. I knew that the general solution of $sin x=sin a$ is $x=npi+(-1)^na$.
$endgroup$
– MrAP
2 days ago
$begingroup$
@MrAP. Consider that two supplementary angles have the same sine. So $$sin(pi- alpha) = sin alpha.$$
$endgroup$
– Matteo
2 days ago
1
$begingroup$
Oh. I got it. Mine generalizes both odd and even n in one equation.
$endgroup$
– MrAP
2 days ago
$begingroup$
@MrAP. Yep. I just prefer to use basic definitions instead of formulae. More ductile, I believe.
$endgroup$
– Matteo
2 days ago
add a comment |
$begingroup$
How did you get the second and third equation from the first equation in your answer. I knew that the general solution of $sin x=sin a$ is $x=npi+(-1)^na$.
$endgroup$
– MrAP
2 days ago
$begingroup$
@MrAP. Consider that two supplementary angles have the same sine. So $$sin(pi- alpha) = sin alpha.$$
$endgroup$
– Matteo
2 days ago
1
$begingroup$
Oh. I got it. Mine generalizes both odd and even n in one equation.
$endgroup$
– MrAP
2 days ago
$begingroup$
@MrAP. Yep. I just prefer to use basic definitions instead of formulae. More ductile, I believe.
$endgroup$
– Matteo
2 days ago
$begingroup$
How did you get the second and third equation from the first equation in your answer. I knew that the general solution of $sin x=sin a$ is $x=npi+(-1)^na$.
$endgroup$
– MrAP
2 days ago
$begingroup$
How did you get the second and third equation from the first equation in your answer. I knew that the general solution of $sin x=sin a$ is $x=npi+(-1)^na$.
$endgroup$
– MrAP
2 days ago
$begingroup$
@MrAP. Consider that two supplementary angles have the same sine. So $$sin(pi- alpha) = sin alpha.$$
$endgroup$
– Matteo
2 days ago
$begingroup$
@MrAP. Consider that two supplementary angles have the same sine. So $$sin(pi- alpha) = sin alpha.$$
$endgroup$
– Matteo
2 days ago
1
1
$begingroup$
Oh. I got it. Mine generalizes both odd and even n in one equation.
$endgroup$
– MrAP
2 days ago
$begingroup$
Oh. I got it. Mine generalizes both odd and even n in one equation.
$endgroup$
– MrAP
2 days ago
$begingroup$
@MrAP. Yep. I just prefer to use basic definitions instead of formulae. More ductile, I believe.
$endgroup$
– Matteo
2 days ago
$begingroup$
@MrAP. Yep. I just prefer to use basic definitions instead of formulae. More ductile, I believe.
$endgroup$
– Matteo
2 days ago
add a comment |
$begingroup$
Use $$sin15^circ=fracsqrt3-12sqrt2$$ and
$$cos15^circ=fracsqrt3+12sqrt2.$$
We obtain:
$$sin(15^circ+x)=sin2x.$$
Thus, $$15^circ+x=2x+360^circk,$$ where $k$ is an integer number, or
$$15^circ+x=180^circ-2x+360^circk.$$
Can you end it now?
answered 2 days ago
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
I had got that equation at the end. Then I equated the angles to get $x=15^circ$
$endgroup$
– MrAP
2 days ago
$begingroup$
@MrAP I got also $55^circ.$
$endgroup$
– Michael Rozenberg
2 days ago
$begingroup$
How did you get that value?
$endgroup$
– MrAP
2 days ago
$begingroup$
@MrAP I added something. See now.
$endgroup$
– Michael Rozenberg
2 days ago
add a comment |
$begingroup$
Use $$sin15^circ=fracsqrt3-12sqrt2$$ and
$$cos15^circ=fracsqrt3+12sqrt2.$$
We obtain:
$$sin(15^circ+x)=sin2x.$$
Thus, $$15^circ+x=2x+360^circk,$$ where $k$ is an integer number, or
$$15^circ+x=180^circ-2x+360^circk.$$
Can you end it now?
answered 2 days ago
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
I had got that equation at the end. Then I equated the angles to get $x=15^circ$
$endgroup$
– MrAP
2 days ago
$begingroup$
@MrAP I got also $55^circ.$
$endgroup$
– Michael Rozenberg
2 days ago
$begingroup$
How did you get that value?
$endgroup$
– MrAP
2 days ago
$begingroup$
@MrAP I added something. See now.
$endgroup$
– Michael Rozenberg
2 days ago
add a comment |
$begingroup$
Use $$sin15^circ=fracsqrt3-12sqrt2$$ and
$$cos15^circ=fracsqrt3+12sqrt2.$$
We obtain:
$$sin(15^circ+x)=sin2x.$$
Thus, $$15^circ+x=2x+360^circk,$$ where $k$ is an integer number, or
$$15^circ+x=180^circ-2x+360^circk.$$
Can you end it now?
answered 2 days ago
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
Use $$sin15^circ=fracsqrt3-12sqrt2$$ and
$$cos15^circ=fracsqrt3+12sqrt2.$$
We obtain:
$$sin(15^circ+x)=sin2x.$$
Thus, $$15^circ+x=2x+360^circk,$$ where $k$ is an integer number, or
$$15^circ+x=180^circ-2x+360^circk.$$
Can you end it now?
answered 2 days ago
Michael RozenbergMichael Rozenberg
108k1895200
edited 2 days ago
answered 2 days ago
Michael RozenbergMichael Rozenberg
108k1895200
answered 2 days ago
Michael RozenbergMichael Rozenberg
108k1895200
answered 2 days ago
Michael RozenbergMichael Rozenberg
108k1895200
108k1895200
$begingroup$
I had got that equation at the end. Then I equated the angles to get $x=15^circ$
$endgroup$
– MrAP
2 days ago
$begingroup$
@MrAP I got also $55^circ.$
$endgroup$
– Michael Rozenberg
2 days ago
$begingroup$
How did you get that value?
$endgroup$
– MrAP
2 days ago
$begingroup$
@MrAP I added something. See now.
$endgroup$
– Michael Rozenberg
2 days ago
add a comment |
$begingroup$
I had got that equation at the end. Then I equated the angles to get $x=15^circ$
$endgroup$
– MrAP
2 days ago
$begingroup$
@MrAP I got also $55^circ.$
$endgroup$
– Michael Rozenberg
2 days ago
$begingroup$
How did you get that value?
$endgroup$
– MrAP
2 days ago
$begingroup$
@MrAP I added something. See now.
$endgroup$
– Michael Rozenberg
2 days ago
$begingroup$
I had got that equation at the end. Then I equated the angles to get $x=15^circ$
$endgroup$
– MrAP
2 days ago
$begingroup$
I had got that equation at the end. Then I equated the angles to get $x=15^circ$
$endgroup$
– MrAP
2 days ago
$begingroup$
@MrAP I got also $55^circ.$
$endgroup$
– Michael Rozenberg
2 days ago
$begingroup$
@MrAP I got also $55^circ.$
$endgroup$
– Michael Rozenberg
2 days ago
$begingroup$
How did you get that value?
$endgroup$
– MrAP
2 days ago
$begingroup$
How did you get that value?
$endgroup$
– MrAP
2 days ago
$begingroup$
@MrAP I added something. See now.
$endgroup$
– Michael Rozenberg
2 days ago
$begingroup$
@MrAP I added something. See now.
$endgroup$
– Michael Rozenberg
2 days ago
add a comment |
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5loN2mHbAVeSR AVI4,qs,ycE,dMq5ouJ0D 5nw Vd7GjKMUvoaSQP,G7bCHDc nf
1
$begingroup$
How did you find the one value that you found? (Perhaps show your working -- it's possible that you accidentally lost some solutions.)
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
I expressed the first numerator in terms of sine and the second in terms of cosine function. Then I used addition formula to combine both sines and cosines. After simplifying, I got $x=pi/12$
$endgroup$
– MrAP
2 days ago
1
$begingroup$
Maybe you accidentally missed some solutions when you got to the simplified stage. Remember that if $sin a = sin b$, then all the solutions (plural!) or $a$ in terms of $b$ are $a = b + 2npi$ or $a = pi - b + 2npi$, for some integer $n$. You will want to take only those solutions for which $x$ comes out to be between $0$ and $pi/2$.
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
@MinusOne-Twelfth, Check my question now.
$endgroup$
– MrAP
2 days ago
1
$begingroup$
You have simply equated the angles. As I said in my previous comment, there are more solutions to $sin a = sin b$ than just $a=b$. This is why you were unable to obtain all the solutions.
$endgroup$
– Minus One-Twelfth
2 days ago