We have $f(x)=\sin 2 x$ in $[0,2 \pi], f^{\prime}(x)=2 \cos 2 x$ For maxima and minima $\mathrm{f}^{\prime}(\mathrm{x})=0 \Rightarrow \cos 2 \mathrm{x}=0$ $\Rightarrow 2 x=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2} \Rightarrow x=\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}$ Now, we find $f(x)$ at $x$
$\begin{aligned}
&\quad=0, \frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}, 2 \pi, \mathrm{f}(0)=0 \\
&\mathrm{f}(\pi / 4)=\sin \pi / 2=1 \mathrm{f}(3 \pi / 4)=\sin 3 \pi / 2 \\
&=-1, \mathrm{f}(5 \pi / 4) \\
&=\sin 5 \pi / 2=1, \mathrm{f}(7 \pi / 4)=\sin 7 \pi / 2 \\
&=-1 \text { and } \mathrm{f}(2 \pi)=\sin 2 \pi=0 \\
&\text { Hence maxima value of } \mathrm{f}(\mathrm{x})=1 \text { at } \mathrm{x}=\pi / 4,5 \pi / 4
\end{aligned}$