Answer
The maximum is $ f(\pi/2) =4$ and the minimum is $f( 3\pi/2) =-4$
Work Step by Step
Given $$y=4 \sin ^{3} \theta-3 \cos ^{2} \theta, \quad[0,2 \pi]$$
Since
\begin{align*}
\frac{dy}{d\theta}&=12 \sin ^{2} \theta \cos \theta+6 \cos \theta \sin \theta
\end{align*}
To find critical points
\begin{align*}
f'(\theta)&=0\\
12 \sin ^{2} \theta \cos \theta+6 \cos \theta \sin \theta&=0\\
(2 \sin \theta +1)6 \cos \theta \sin \theta &=0
\end{align*}
Then $\theta \in [0,2\pi]$ $$\theta=0, \quad \pi / 2, \quad \pi, \quad 7 \pi / 6, \quad 3 \pi / 2, \quad 11 \pi / 6, \quad 2 \pi$$
Since
\begin{align*}
f(0)&=-3 \\
f(\pi/2)&=4\\
f(\pi)&=-3\\
f(7\pi/6)&=-11/4\\
f(3\pi/2)&=-4\\
f(11\pi/6)&=-11/4\\
f(2\pi)&=-3
\end{align*}
Then the maximum is $ f(\pi/2) =4$ and the minimum is $f( 3\pi/2) =-4$
