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Trigonometric equations - Find the general solution

  1. Jul 14, 2013 #1
    1. The problem statement, all variables and given/known data
    [tex]\cos x-\sqrt{3}\sin x=\cos(3x)[/tex]


    2. Relevant equations



    3. 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!
     
  2. jcsd
  3. Jul 14, 2013 #2

    ehild

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

    ehild
     
  4. Jul 14, 2013 #3
    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.
     
  5. Jul 14, 2013 #4

    LCKurtz

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    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: Jul 14, 2013
  6. Jul 14, 2013 #5

    verty

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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).
     
  7. Jul 14, 2013 #6

    haruspex

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    Yes, both answers reduce to ##n\pi, n\pi+\pi/6, n\pi+\pi/3##
     
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