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7 November, 06:13

Which values of x in the interval 0 x 360 satisfy the equation 2sin^2x+sinx-1=0?

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  1. 7 November, 06:36
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    Factor first: (sinx+1) (2sinx-1) = 0

    sinx+1=0 or 2sinx-1=0

    sinx=-1 or sinx=1/2

    0≤x≤360, so x=270, or x=30 or x=150
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