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A problem in finding the General Solution of a Trigonometric Equation

  1. May 18, 2017 #1
    1. The problem statement, all variables and given/known data:

    Find the general solution of the Trigonometric equation $$\sin {3x}+\sin {x}=\cos {6x}+\cos {4x} $$

    Answers given are: ##(2n+1)\frac {\pi}{2}##, ##(4n+1)\frac {\pi}{14}## and ##(4n-1)\frac {\pi}{6}##.

    2. Relevant equations:

    Equations that may be used:

    20170519_023122.png

    3. The attempt at a solution:

    Please see the attached pic:

    1495140875722-1985572902.jpg

    The answer from Case 1 is correct, but I can't find my mistake in the answers from the two sub-cases of case 2.
     
  2. jcsd
  3. May 18, 2017 #2

    Charles Link

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    Homework Helper

    Your mistake is in the trigonometric identity ## cosA-cosB=-2sin((A+B)/2)sin((A-B)/2) ##. You didn't divide the terms (A+B and A-B) by 2 in both cases.
     
  4. May 18, 2017 #3
    Got it. Thanks a lot.
     
  5. May 18, 2017 #4

    Charles Link

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    Homework Helper

    @Wrichik Basu This is an extra detail, but it may interest you that I think the solution ## x=(2n+1) \frac{\pi}{2} ## for all integers ## n ## is actually all included in the other two solutions. The reason is that ## x=(2n+1) \frac{\pi}{2} ## is also always a solution of ## cos(5x)=sin(2x) ##. (A complete expansion of ## cos(5x) ## and ## sin(2x) ##will generate a ## cos(x) ## factor on both sides of the equation.) ## \\ ## You can write ## x= (2k+1) \frac{\pi}{2}=(4n+1) \frac{\pi}{14} ## and if ## k ## is odd, for any ## k ## you can find an integer ## n ##. You can also write ## x=(2k+1) \frac{\pi}{2}=(4m-1) \frac{\pi}{6} ## and if ## k ## is even, for any ## k ## you can find an integer ## m ##. Thereby, the last two solutions completely overlap the ## x=(2n+1) \frac{\pi}{2} ## solution. ## \\ ## Editing... The other two solutions are completely independent of each other=a little algebra shows there is no "x" that is the same in both of them.
     
    Last edited: May 18, 2017
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