TonnyB6420 TonnyB6420

16-01-2018
Mathematics

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

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tynollette tynollette

16-01-2018

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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