Answer
The solutions are $\frac{\pi}{2}$ and $\frac{3\pi}{2}$.
Work Step by Step
Take the square root of both sides to obtain:
$\sin^2{x} = 1
\\\sqrt{\sin^2{x}}=\pm\sqrt1
\\\sin{x} = \pm 1$
Thus, $\sin{x} = -1$ or $\sin{x} = 1$.
Solve each equation to obtain:
\begin{array}{ccc}
&\sin{x}=-1 &\text{or} &\sin{x} = 1
\\&x=\sin^{-1}{(-1)} &\text{or} &x=\sin^{-1}{(1)}
\\&x=-\frac{\pi}{2} &\text{or} &x=\frac{\pi}{2}
\end{array}
Only $\frac{\pi}{2}$ is in the interval $[0, 2\pi)$.
However, since the sine function is sinusoidal, note that $\sin{(\frac{3\pi}{2})}=-1$, which makes $\sin^2{(\frac{3\pi}{2})}=1$.
Therefore, the solutions are $\frac{\pi}{2}$ and $\frac{3\pi}{2}$.
