At what points in the interval [0, 2π],

Let $f(x)=\sin 2 x$.

$\therefore f^{\prime}(x)=2 \cos 2 x$

Now,

$f^{\prime}(x)=0 \Rightarrow \cos 2 x=0$

$\Rightarrow 2 x=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2}$

$\Rightarrow x=\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}$

Then, we evaluate the values of $f$ at critical points $x=\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}$ and at the end points of the interval $[0,2 \pi]$.

$f\left(\frac{\pi}{4}\right)=\sin \frac{\pi}{2}=1, f\left(\frac{3 \pi}{4}\right)=\sin \frac{3 \pi}{2}=-1$

$f\left(\frac{5 \pi}{4}\right)=\sin \frac{5 \pi}{2}=1, f\left(\frac{7 \pi}{4}\right)=\sin \frac{7 \pi}{2}=-1$

$f(0)=\sin 0=0, f(2 \pi)=\sin 2 \pi=0$

Hence, we can conclude that the absolute maximum value of $f$ on $[0,2 \pi]$ is occurring at $x=\frac{\pi}{4}$ and $x=\frac{5 \pi}{4}$.