Answer
$\theta= \dfrac{3\pi}{4}+n\pi $
$ g(\theta)$ has neither a local maximum nor a local minimum.
Work Step by Step
Given $$g(\theta)=\sin ^{2} \theta+\theta$$
Since
\begin{align*}
g'(\theta) &=2\sin \left(θ\right)\cos \left(θ\right)+1 =\sin 2\theta +1\\
g''(\theta) &= 2\cos 2\theta
\end{align*}
Since $g(\theta)$ has critical points when $g'(\theta)= 0$
\begin{align*}
g'(\theta)&=0\\
\sin 2\theta +1&=0 \\
\sin 2\theta&=-1
\end{align*}
Then $g(\theta)$ has critical points at $\theta= \dfrac{3\pi}{4}+n\pi $. We note that
\begin{align*}
g'(\theta)&\gt0
\end{align*}
For all $\theta$. Thus, $ g(\theta)$ has neither a local maximum nor a local minimum.