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Math Help - solving trig functions sin2x

  1. #1
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    solving trig functions sin2x

    This is the question

    sin2x-0.8=0 in the interval 2x [0,2pi]

    so far i got this
    sin2x=0.8
    2x=sin inverse 0.8
    2x=0.927
    x=0.46

    the book has two answer 0.46, and 1.11. How did the book got 1.11?
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  2. #2
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    Quote Originally Posted by jepal View Post
    This is the question

    sin2x-0.8=0 in the interval 2x [0,2pi]

    so far i got this
    sin2x=0.8
    2x=sin inverse 0.8
    2x=0.927
    x=0.46

    the book has two answer 0.46, and 1.11. How did the book got 1.11?
    This is the solution that lies in the second quadrant. You get it via 2x = pi - 0.927.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    This is the solution that lies in the second quadrant. You get it via 2x = pi - 0.927.
    will this be the case too if it was cos2x.?
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  4. #4
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    Quote Originally Posted by jepal View Post
    will this be the case too if it was cos2x.?
    If it was cos(2x) = 0.8 then you would need to remember that 2x is in the first quadrant or the fourth quadrant ....
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