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Math Help - [SOLVED] Correct double angle working?

  1. #1
    Senior Member Stroodle's Avatar
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    367

    [SOLVED] Correct double angle working?

    The question asks to solve the following for x

    cos(2x)cos(\frac{\pi}{6})-sin(2x)sin(\frac{\pi}{6})=\frac{1}{2} where x\in\left [0,2\pi\right ]

    My working:

    cos(2x)cos(\frac{\pi}{6})-sin(2x)sin(\frac{\pi}{6})=cos(2x+\frac{\pi}{6})

    \Rightarrow cos(2x+\frac{\pi}{6})=\frac{1}{2},\ x\in\left [0,2\pi\right ]

    2x+\frac{\pi}{6}=\frac{\pi}{3},\ 2\pi-\frac{\pi}{3},\ 2\pi+\frac{\pi}{3},\ 4\pi-\frac{\pi}{3}

    x=\frac{\pi}{12}, x=\frac{3\pi}{4}, x=\frac{7\pi}{4} and x=\frac{13\pi}{12}

    I've been told it's wrong, with no explanation as to why.... As usual, I'm probably missing something really obvious?

    Thanks for your help!


    * Edit - nvm, just had it rechecked and apparently my working is correct...
    Last edited by Stroodle; April 8th 2010 at 10:28 PM. Reason: added comment
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  2. #2
    MHF Contributor Amer's Avatar
    Joined
    May 2009
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    Jordan
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    \cos (2x+\frac{\pi}{6} ) = \frac{1}{2}

    you know that
    \cos x = \frac{1}{2}

    x = \frac{\pi}{3} + 2k\pi ...(1)

    x= \frac{-\pi}{3} + 2k\pi ...(2)

    so

    2x + \frac{\pi}{6} = \frac{\pi}{3} + 2k\pi

    x = \frac{\pi}{12} + k\pi

    but x \in \left[0, \pi\right]

    so x = \frac{\pi}{12} , \frac{\pi}{12} + \pi
    this form the first one

    leave the rest for you
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