Answer
The maximum is $g(2\pi/3)= g( 4\pi/3)=5/4$, the minimum is $g(0)=g(2\pi )=-1 $
Work Step by Step
Given $$g(\theta)=\sin ^{2} \theta-\cos \theta, \quad[0,2 \pi]$$
Since
$$ g'(\theta)= 2\sin \theta\cos \theta+\sin \theta$$
We find the critical points
\begin{align*}
g'(\theta)&=0\\
2\sin \theta\cos \theta+\sin \theta&=0\\
\sin \theta(2\cos \theta+1)&=0
\end{align*}
Then $g(\theta)$ has critical points at $$ \theta=0, \frac{2 \pi}{3}, \pi, \frac{4 \pi}{3}, 2 \pi$$
We check the values at the critical points and end points:
\begin{align*}
g(0)&= -1 \\
g(2\pi/3)&=5/4 \\
g(\pi)&= 1 \\
g(4\pi/3)&= 5/4 \\
g(2\pi)&= -1
\end{align*}
Then the maximum is $g(2\pi/3)= g( 4\pi/3)=5/4$, and the minimum is $g(0)=g(2\pi )=-1 $
