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Math Help - trigo equation for Sinus x,3x,5x

  1. #1
    Super Member dhiab's Avatar
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    trigo equation for Sinus x,3x,5x

    Solve in this equation :

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  2. #2
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    Grandad's Avatar
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    Hello dhiab
    Quote Originally Posted by dhiab View Post
    Solve in this equation :

    Using 2\sin^2\theta = 1 - \cos2\theta, the equation can be re-written as:

    1 - \cos2x + 1 - \cos6x +1-\cos10x = 3

    \Rightarrow \cos2x+\cos6x + \cos10x = 0

    Now use \cos A + \cos B = \cos\tfrac12(A+B)\cos\tfrac12(A-B) to write \cos2x + \cos10x as \cos6x\cos4x. So:

    \cos6x(1+\cos4x)=0

    \Rightarrow \cos6x = 0 or \cos4x = -1

    \Rightarrow 6x = \tfrac{\pi}{2}, \tfrac{3\pi}{2}, \tfrac{5\pi}{2}, \tfrac{7\pi}{2}, \tfrac{9\pi}{2}, \tfrac{11\pi}{2}

    or 4x = \pi, 3\pi

    \Rightarrow x = \tfrac{\pi}{12}, \tfrac{\pi}{4}, \tfrac{5\pi}{12}, \tfrac{7\pi}{12}, \tfrac{3\pi}{4}, \tfrac{11\pi}{12}

    Grandad
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  3. #3
    Super Member dhiab's Avatar
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    Quote Originally Posted by Grandad View Post
    Hello dhiabUsing 2\sin^2\theta = 1 - \cos2\theta, the equation can be re-written as:

    1 - \cos2x + 1 - \cos6x +1-\cos10x = 3

    \Rightarrow \cos2x+\cos6x + \cos10x = 0

    Now use \cos A + \cos B = \cos\tfrac12(A+B)\cos\tfrac12(A-B) to write \cos2x + \cos10x as \cos6x\cos4x. So:

    \cos6x(1+\cos4x)=0

    \Rightarrow \cos6x = 0 or \cos4x = -1

    \Rightarrow 6x = \tfrac{\pi}{2}, \tfrac{3\pi}{2}, \tfrac{5\pi}{2}, \tfrac{7\pi}{2}, \tfrac{9\pi}{2}, \tfrac{11\pi}{2}

    or 4x = \pi, 3\pi

    \Rightarrow x = \tfrac{\pi}{12}, \tfrac{\pi}{4}, \tfrac{5\pi}{12}, \tfrac{7\pi}{12}, \tfrac{3\pi}{4}, \tfrac{11\pi}{12}

    Grandad
    [B]Hello : Thank you
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