Answer
$ \dfrac{\pi}{6} + 2 \pi k, \, \dfrac{5\pi}{6} + 2 \pi k,\, 2 \pi k - \dfrac{\pi}{2}$
Work Step by Step
$2-2 \sin^2{x} - \sin{x} = 1$
$2 \sin^2{x} + \sin{x} - 1 = 0$
Using the quadratic formula:
$\sin{x} = \dfrac{-1 \pm \sqrt{1-4(2)(-1)}}{2(2)} = \dfrac{-1 \pm 3}{4}$
$\sin{x} = \dfrac{1}{2} \hspace{30pt} \sin{x} = -1 $
$x = \dfrac{\pi}{6} + 2 \pi k, \, \dfrac{5\pi}{6} + 2 \pi k,\, 2 \pi k - \dfrac{\pi}{2}$
