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Math Help - How can I approximately solve trigonometric equations simultaneously?

  1. #1
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    How can I approximately solve trigonometric equations simultaneously?

    if I have p/k=sin(3x)/sin(2x), how could I solve for x in terms of p/k? I understand that these are not linear functions so they cannot be solved simultaneously but is it possible to approximately solve this problem using taylor expansions? If there is another way I could solve this, I would like to hear what you think. Thank you
    Last edited by Marckutt; February 11th 2011 at 06:32 PM.
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  2. #2
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    You need to use Trigonometric Identities...

    \displaystyle \sin{3x} \equiv \sin{x}(3 - 4\sin^2{x}), \sin{2x} \equiv 2\sin{x}\cos{x} and \displaystyle \sin^2{x} + \cos^2{x} = 1.


    So if \displaystyle \frac{\sin{3x}}{\sin{2x}} = \frac{p}{k}

    \displaystyle \frac{\sin{x}(3-4\sin^2{x})}{2\sin{x}\cos{x}} = \frac{p}{k}

    \displaystyle \frac{3 - 4\sin^2{x}}{2\cos{x}} = \frac{p}{k}

    \displaystyle \frac{3 - 4(1 - \cos^2{x})}{\cos{x}} = \frac{p}{k}

    \displaystyle \frac{3 - 4 + 4\cos^2{x}}{\cos{x}} = \frac{p}{k}

    \displaystyle \frac{4\cos^2{x} - 1}{\cos{x}} = \frac{p}{k}

    \displaystyle 4\cos^2{x} - 1 = \frac{p}{k}\cos{x}

    \displaystyle 4\cos^2{x} - \frac{p}{k}\cos{x} - 1 = 0.


    Now let \displaystyle X = \cos{x} and the equation becomes

    \displaystyle 4X^2 - \frac{p}{k}X - 1 = 0,

    which is a quadratic equation that you can solve for \displaystyle X, the solution of which you can use to solve for \displaystyle x.
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  3. #3
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    thank you very much
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