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Math Help - How do I do this?

  1. #1
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    How do I do this?

    I have to find all solutions of the equation in the interval [0, 2pi) for:
    sin 2x = (- sqrt3 / 2)

    My work:
    sin 2x = -sqrt(3)/2
    2x = 5pi/3, 4pi/3
    x = 5pi/6 or 2pi/3

    Is this correct?
    Last edited by iluvmathbutitshard; November 24th 2009 at 08:17 AM.
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  2. #2
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    Quote Originally Posted by iluvmathbutitshard View Post
    I have to find all solutions of the equation in the interval [0, 2pi) for:
    sin 2x = (- sqrt3 / 2)

    I am not sure to do this.
    Any help is appreciated.
    0 \le x < 2\pi

    0 \le 2x < 4\pi

    values in the interval [0, 4\pi) where

    \sin(2x) = -\frac{\sqrt{3}}{2} ...

    2x = \frac{4\pi}{3} \, , \, \frac{5\pi}{3} \, , \, \frac{10\pi}{3} \, , \, \frac{11\pi}{3}

    solve for x
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  3. #3
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    so:
    x = 5pi/6, 4pi/6, 10pi/6, 11pi/6

    Is this right?
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  4. #4
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    Quote Originally Posted by iluvmathbutitshard View Post
    so:
    x = 5pi/6, 4pi/6, 10pi/6, 11pi/6

    Is this right?
    reduce the fractions that can be reduced.
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