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Math Help - tough inverse function problem

  1. #1
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    tough inverse function problem

    1.) -8sin x/2= -4 square root 3

    i have no idea how to even start it???
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  2. #2
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    divide both sides by -8

    this should give you something that looks workable.
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  3. #3
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    do you know how to start sin2x= -1/2?
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  4. #4
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     \sin(2x) = \frac{-1}{2}

    what domain does this need to be solved over?

    making 2x = u

     \sin(u) = \frac{-1}{2}

    this gives

     u = \pi+\frac{\pi}{6}, 2\pi-\frac{\pi}{6} = \frac{7\pi}{6}, \frac{11\pi}{6}

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

     x = \frac{7\pi}{12}, \frac{11\pi}{12}
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