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Math Help - Solving for x

  1. #1
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    Solving for x

    This is part of a bigger question to do with polar coordinates - I have obtained a derivative and equated it to zero in order to indicate a stationary point and this is what I need to solve:

    2cos(2x)cos(x) - sin(2x)sin(x) = 0

    Which of the double/half/compound angle formulas do I need to use here? I always seem to forget in questions like these.

    Thanks if you can help me :]
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  2. #2
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    Quote Originally Posted by db5vry View Post
    This is part of a bigger question to do with polar coordinates - I have obtained a derivative and equated it to zero in order to indicate a stationary point and this is what I need to solve:

    2cos(2x)cos(x) - sin(2x)sin(x) = 0

    Which of the double/half/compound angle formulas do I need to use here? I always seem to forget in questions like these.

    Thanks if you can help me :]
    Dear db5vry,

    Try, sin2x=2sinxcosx~and~cos2x=2cos^{2}x-1=1-2sin^{2}x=cos^{2}x-sin^{2}x

    Hoe this will help you.
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  3. #3
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    Quote Originally Posted by Sudharaka View Post
    Dear db5vry,

    Try, sin2x=2sinxcosx~and~cos2x=2cos^{2}x-1=1-2sin^{2}x=cos^{2}x-sin^{2}x

    Hoe this will help you.
    Will this then lead me to

    cosx (6sin^2x - 2) = 0 which gives the solutions \frac{\pi}{2} and \frac{\sqrt{3}}{3}? If so then I've done this correctly, but I'm not sure!
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  4. #4
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    Quote Originally Posted by db5vry View Post
    Will this then lead me to

    cosx (6sin^2x - 2) = 0 which gives the solutions \frac{\pi}{2} and \frac{\sqrt{3}}{3}? If so then I've done this correctly, but I'm not sure!
    Dear db5vry,

    You could eliminate your doubts by substituting the solutions in the original equation.
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