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Math Help - cos 3x sin2x = 1 /16 (2 cos x - cos 3x - cos 5x)

  1. #1
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    cos 3x sin2x = 1 /16 (2 cos x - cos 3x - cos 5x)

    Please solve cos 3x sin2x = 1 /16 (2 cos x - cos 3x - cos 5x)
    Last edited by mr fantastic; April 27th 2009 at 02:01 AM. Reason: Deleted excessive questions marks
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  2. #2
    MHF Contributor red_dog's Avatar
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    16\cos 3x\sin 2x=(\cos x-\cos 3x)+(\cos x-\cos 5x)

    16\cos 3x\sin 2x=2\sin x\sin 2x+2\sin 2x\sin 3x

    \sin 2x(8\cos 3x-\sin x-\sin 3x)=0

    \sin 2x=0\Rightarrow 2x=k\pi\Rightarrow x=\frac{k\pi}{2}, \ k\in\mathbf{Z}

    8\cos 3x-(\sin x+\sin 3x)=0

    8\cos 3x-2\sin 2x\cos x=0

    4(4\cos^3x-3\cos x)-\sin 2x\cos x=0

    \cos x(16\cos^2x-12-\sin 2x)=0

    \cos x=0\Rightarrow x=\frac{(2k+1)\pi}{2}, \ k\in\mathbf{Z}

    16\cos^2x-12-2\sin x\cos x=0

    16\cos^2x-12(\sin^2x+\cos^2x)-2\sin x\cos x=0

    2\cos^2x-6\sin^2x-\sin x\cos x=0

    Divide by \sin^2x

    2-6\tan^2x-\tan x=0

    Substitute \tan x=t and solve the quadratic.
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