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Thread: help, please and thank you

  1. #1
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    Exclamation help, please and thank you

    find exact solutions of 2sin2x = -1, for 0 (<or=)x<2pi

    Thanks a bunch
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  2. #2
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    Quote Originally Posted by Luckegrl_16 View Post
    find exact solutions of 2sin2x = -1, for 0 (<or=)x<2pi

    Thanks a bunch
    $\displaystyle 2sin(2x) = -1$

    $\displaystyle sin(2x) = -\frac{1}{2}$

    $\displaystyle 2x = \frac{7\pi}{6}, \frac{11\pi}{6}$

    So
    $\displaystyle x = \frac{7\pi}{12}, \frac{11\pi}{12}$

    -Dan
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  3. #3
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    yaya

    oh and srry about the whole multiple posts thing, I've never really used these b4, but now I know
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