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Trigonometric equation

  1. Sep 2, 2016 #1
    1. The problem statement, all variables and given/known data
    The following equation is to be solved for all x:
    ## \cos(x) + \cos(3x) = \sin(x) + \sin(3x)##

    2. Relevant equations
    The tripple angle formulas:
    ## \cos(3x) = 4\cos^3(x) - 3\cos(x) ##
    ##\sin(3x) = 3\sin(x) - 4\sin^3(x) ##
    The Pythagorean trig identity:
    ## \sin^2(x) + \cos^2(x) = 1 ##

    3. The attempt at a solution
    Applying the tripple angle identities we have:
    ## \cos(x) + 4\cos^3(x) - 3\cos(x) = \sin(x) + 3\sin(x) - 4\sin^3(x) ##
    Simplifying:
    ## 4\cos^3(x) - 2\cos(x) = 4\sin(x) - 4\sin^3(x) ##
    ## 2\cos(x)(2\cos^2(x) - 1) = 4\sin(x)(1 - \sin^2(x) ##
    With the Pythagorean identity:
    ## \cos(x)(\cos^2(x) - \sin^2(x)) = 2\sin(x)\cos^2(x)##

    Now, from this it look as if ##cos(x) = 0## should be a solution, which yields ##x_1 = \pi n## where ##n## is an integer. Continuing with the rest:

    ## (\cos^2(x) - \sin^2(x)) = 2\sin(x)\cos(x)##
    ## \cot(x) - \tan(x) = 2 ##
    ## \tan^2(x) + 2\tan(x) -1 =0##

    Solving the second degree equation yields ##\tan(x) = -1 \pm \sqrt{2} ##, which gives ##x_2 = \frac{\pi}{8} +\pi n## and ##x_3 = \frac{-3\pi}{8} +\pi n##.

    Now the only problem is that none of these solutions is right! Here https://www.desmos.com/calculator/dvbz4qpadt I plotted the functions and searched their intersection, and it doesn't match my solution. Where is my misstake? How can I solve the equation properly?

    Thank you very much in advance!
     
  2. jcsd
  3. Sep 2, 2016 #2

    micromass

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    Are you sure that ##\cos(n\pi) = 0##?
     
  4. Sep 2, 2016 #3
    Oh noo! That's totally wrong! It yields ##x_1 = \frac{\pi}{2} + \pi n## of course!
     
  5. Sep 2, 2016 #4

    micromass

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    Yep. And now it seems the graph agrees with your solutions!
     
  6. Sep 2, 2016 #5
    And now I see that the solutions are fine...Such a stupid misstake!
    Thank you for your help!
     
  7. Sep 2, 2016 #6

    haruspex

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    Fwiw, you could have gone
    ##\cos(2x-x)+\cos(2x+x)=\sin(2x-x)+\sin(2x+x)##
    ##2\cos(2x)\cos(x)=2\sin(2x)\cos(x)##
    ##\cos(x)=0## or ##\tan(2x)=1##.
     
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