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Math Help - trigonometric equation

  1. #1
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    trigonometric equation

    solve for x
    4(\cos^3 x + \sin^3 x) = \cos x + 3 \sin x
    Last edited by mr fantastic; July 21st 2009 at 09:48 PM. Reason: Corrected equation
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  2. #2
    MHF Contributor red_dog's Avatar
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    It is well known that \sin^2x+\cos^2x=1. Using this the equation can be written as

    4(\cos^3x+\sin^3x)=\cos x(\sin^2x+\cos^2x)+3\sin x(\sin^2x+\cos^2x)

    \sin^3x-\sin^2x\cos x-3\sin x\cos^2x+3\cos^2x=0

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

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

    Now, \sin x-\cos x=0\Rightarrow\tan x=1\Rightarrow x=\frac{\pi}{4}+k\pi, \ k\in\mathbb{Z}

    or \sin^2x-3\cos^2x=0\Rightarrow\tan^2x=3\Rightarrow \tan x=\pm\sqrt{3}\Rightarrow x=\pm\frac{\pi}{3}+k\pi, \ k\in\mathbb{Z}
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  3. #3
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    good, thanks red dog much.
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