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Trigonometric equations - Find the general solution

  • Thread starter Saitama
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  • #1
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Homework Statement


[tex]\cos x-\sqrt{3}\sin x=\cos(3x)[/tex]


Homework Equations





The Attempt at a Solution


Dividing both the sides by two i.e
[tex]\cos x \cos \frac{\pi}{3}-\sin x \sin \frac{\pi}{3}=\cos (3x)/2[/tex]
LHS can be written as ##\cos(x+\pi/3)##. Substituting ##x+\pi/3=t \Rightarrow 3x=3t-\pi \Rightarrow \cos(3x)=-\cos(3t)##
[tex]\cos t=-\cos(3t)/2 \Rightarrow 2\cos t+\cos(3t)=0[/tex]
##\because \cos(3t)=4\cos^3t-3\cos t##
[tex]4\cos^3t-\cos t=0 \Rightarrow \cos t(4\cos^2t-1)=0[/tex]
Hence, ##\cos t=0 \Rightarrow t=(2n+1)\pi/2 \Rightarrow x=(2n+1)\pi/2-\pi/3## and ##\cos t=±1/2 \Rightarrow t=k\pi ± \pi/3 \Rightarrow x=k\pi, k\pi-2\pi/3## where ##n,k \in Z##.

But these are wrong. The correct answers are: ##n\pi, \pi(3k+(-1)^k)/6##.

Any help is appreciated. Thanks!
 

Answers and Replies

  • #2
ehild
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It is too complicated. Try to expand cos(3x)=cos(x+2x) in the original equation.

ehild
 
  • #3
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It is too complicated. Try to expand cos(3x)=cos(x+2x) in the original equation.

ehild
I already devised a way to reach the correct answer. What I have posted above is my first attempt at the problem and I am curious as to where I went wrong with that.

To reach the correct answer, I would rewrite the given equation as ##\cos(3x)-\cos x+\sqrt{3}\sin x=0##. Rewriting ##\cos(3x)-\cos x## as ##-2\sin(2x)\sin x##, I can easily reach the correct answer.
 
  • #4
LCKurtz
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Homework Statement


[tex]\cos x-\sqrt{3}\sin x=\cos(3x)[/tex]


Homework Equations





The Attempt at a Solution


Dividing both the sides by two i.e
[tex]\cos x \cos \frac{\pi}{3}-\sin x \sin \frac{\pi}{3}=\cos (3x)/2[/tex]
LHS can be written as ##\cos(x+\pi/3)##. Substituting ##x+\pi/3=t \Rightarrow 3x=3t-\pi \Rightarrow \cos(3x)=-\cos(3t)##
[tex]\cos t=-\cos(3t)/2 \Rightarrow 2\cos t+\cos(3t)=0[/tex]
##\because \color{red}{\cos(3t)=4\cos^3t-3\cos t}##
[tex]4\cos^3t-\cos t=0 \Rightarrow \cos t(4\cos^2t-1)=0[/tex]
Hence, ##\cos t=0 \Rightarrow t=(2n+1)\pi/2 \Rightarrow x=(2n+1)\pi/2-\pi/3## and ##\cos t=±1/2 \Rightarrow t=k\pi ± \pi/3 \Rightarrow x=k\pi, k\pi-2\pi/3## where ##n,k \in Z##.

But these are wrong. The correct answers are: ##n\pi, \pi(3k+(-1)^k)/6##.

Any help is appreciated. Thanks!
Where did that equation I have highlighted in red come from?

[Edit - added] Never mind about the identity. I think your work is correct and your answer and the "correct" answer are identical.
 
Last edited:
  • #5
verty
Homework Helper
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Pranav, I followed exactly your method and got the right answer. Your mistake must be after the word "hence". I got (using degrees) x = 180k + (0,30,60).
 
  • #6
haruspex
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I think your work is correct and your answer and the "correct" answer are identical.
Yes, both answers reduce to ##n\pi, n\pi+\pi/6, n\pi+\pi/3##
 

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