0 <= x <= 2pi (sin^2x)-(2sinx)=3
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Originally Posted by wisezeta 0 <= x <= 2pi (sin^2x)-(2sinx)=3 $\displaystyle \sin^2{x} - 2\sin{x} - 3 = 0$ $\displaystyle (\sin{x}-3)(\sin{x}+1) = 0 $ $\displaystyle x = \frac{3\pi}{2} $
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