Answer
$-\frac{3\pi}{4}, -\frac{\pi}{4},\frac{\pi}{4}, \frac{3\pi}{4}$
Work Step by Step
1. Given $sin^2(s)=\frac{1}{2}$, we have $sin(s)=\pm\frac{\sqrt 2}{2}$.
2. For $sin(s)=\frac{\sqrt 2}{2}$, we can find a reference angle as $s_0=sin^{-1}(\frac{\sqrt 2}{2})=\frac{\pi}{4}$. In $[-\pi,\pi)$, there are two solution angles $s=\frac{\pi}{4}, \pi-\frac{\pi}{4}$ or $s=\frac{\pi}{4}, \frac{3\pi}{4}$
3. For $sin(s)=-\frac{\sqrt 2}{2}$, we can find a reference angle as $s_0=sin^{-1}(\frac{\sqrt 2}{2})=\frac{\pi}{4}$. In $[-\pi,\pi)$, there are two solution angles $s=-\pi+\frac{\pi}{4}, -\frac{\pi}{4}$ or $s=-\frac{3\pi}{4}, -\frac{\pi}{4}$
4. The solutions are $-\frac{3\pi}{4}, -\frac{\pi}{4},\frac{\pi}{4}, \frac{3\pi}{4}$
