Answer
The maximum is $ f(5\pi/3) \approx 6.968$ and the minimum is $f( \pi/ 3) \approx - 0.685$
Work Step by Step
Given $$y=\theta-2 \sin \theta, \quad[0,2 \pi]$$
Since
\begin{align*}
\frac{dy}{d\theta}&=1-2\cos \theta
\end{align*}
To find critical points
\begin{align*}
f'(\theta)&=0\\
1-2\cos \theta&=0\\
\cos \theta &=\frac{1}{2}
\end{align*}
Then $\theta= \pi/3,\ \ 5\pi/3$. Because $\pi/3,\ 5\pi/3\in [0,2\pi]$, then $f(\theta)$ has a critical point $\theta=\pi/3,\ \ 5\pi/3$, since
\begin{align*}
f(0)&\approx 6.283 \\
f(\pi/3)&\approx- 0.685\\
f(5\pi/3)&\approx 6.968\\
f(2\pi/)&\approx 6.283
\end{align*}
Then the maximum is $ f(5\pi/3) \approx 6.968$ and the minimum is $f( \pi/ 3) \approx - 0.685$
