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Math Help - solving for x

  1. September 3rd 2009, 07:16 AM #1
    wisezeta
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    solving for x

    0 <= x <= 2pi

    (sin^2x)-(2sinx)=3
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  2. September 3rd 2009, 07:22 AM #2
    skeeter
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    Quote Originally Posted by wisezeta View Post
    0 <= x <= 2pi

    (sin^2x)-(2sinx)=3
    \sin^2{x} - 2\sin{x} - 3 = 0

    <br /> (\sin{x}-3)(\sin{x}+1) = 0<br />

    <br /> x = \frac{3\pi}{2}<br />
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