Answer
$-\frac{\pi}{2}, \frac{3\pi}{2}, \frac{7\pi}{2}, \frac{11\pi}{2} $
Work Step by Step
Let $2sin(x)=-2\longrightarrow sin(x)=-1$, we can find a general form of the solutions as $x=2k\pi+\frac{3\pi}{2}$ where $k$ is an integer. For $k=-1,0,1,2$, we have $x=-\frac{\pi}{2}, \frac{3\pi}{2}, \frac{7\pi}{2}, \frac{11\pi}{2} $
