Write the values of x in [0, π] for which sin 2x,

(i) $\sin 2 x, \frac{1}{2}$ and $\cos 2 x$ are in AP.

$\therefore \sin 2 x+\cos 2 x=2 \times \frac{1}{2}$

$\Rightarrow \sin 2 x+\cos 2 x=1 \quad \ldots$ (1)

This equation is of the form $a \sin \theta+b \cos \theta=c$, where $a=1, b=1$ and $c=1$.

Now,

Let:

$a=r \sin \alpha$ and $b=r \cos \alpha$

Thus, we have:

$r=\sqrt{a^{2}+b^{2}}=\sqrt{1^{2}+1^{2}}=\sqrt{2}$ and $\tan \alpha=1 \Rightarrow \alpha=\frac{\pi}{4}$

On putting $a=1=r \sin \alpha$ and $b=1=r \cos \alpha$ in equation (1), we get:

$r \sin \alpha \sin 2 x+r \cos \alpha \cos 2 x=1$

$\Rightarrow r \cos (2 x-\alpha)=1$

$\Rightarrow \sqrt{2} \cos \left(2 x-\frac{\pi}{4}\right)=1$

$\Rightarrow \cos \left(2 x-\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$

$\Rightarrow \cos \left(2 x-\frac{\pi}{4}\right)=\cos \frac{\pi}{4}$

$\Rightarrow 2 x-\frac{\pi}{4}=2 n \pi \pm \frac{\pi}{4}, \mathrm{n} \in \mathrm{Z}$

Taking positive value, we get:

$\Rightarrow 2 \mathrm{x}-\frac{\pi}{4}=2 \mathrm{n} \pi+\frac{\pi}{4}$

$\Rightarrow \mathrm{x}=n \pi+\frac{\pi}{4}$

Taking positive value, we get:

$\Rightarrow 2 \mathrm{x}-\frac{\pi}{4}=2 n \pi-\frac{\pi}{4}$

$\Rightarrow 2 \mathrm{x}-\frac{\pi}{4}=2 n \pi-\frac{\pi}{4}$

$\Rightarrow x=n \pi, n \in Z$

For $n=0$, the values of $x$ are $\frac{\pi}{4}$ and 0 and for $n=1$, the values of $x$ are $\frac{5 \pi}{4}$ and $\pi$. $\frac{5 \pi}{4}$ does not satisfy the condition.

For the other value of $n$, the given condition is not true, i.e., $[0, \pi]$.