Answer
$\theta=\frac{n\pi}{2}$
Work Step by Step
Let $\sin \theta =u$
Then, $g(\theta)=\sin^{2}\theta=u^{2}$
$g'(\theta)=\frac{d}{d\theta}(u^{2})=2u\frac{du}{d\theta}=2\sin\theta\frac{d}{d\theta}(\sin\theta)$
$=2\sin\theta \cos\theta$
If $g'(\theta)$ does not exist or $g'(\theta)=0$, this implies $\theta$ is a critical point of the function. But, $g'(\theta)$ exists for all $\theta$.
If $g'(\theta)=2 \sin\theta\cos \theta=0$
$\implies \sin\theta=0$ or $\cos \theta= 0$
$\implies \theta =n\pi$ or $\theta =(2n+1)\frac{\pi}{2}$ where $n$ is any integer.
Combining $\theta=n\pi$ and $\theta=(2n+1)\frac{\pi}{2}$, we can write
$g'(\theta)=0\implies \theta=\frac{n\pi}{2}$
