The values of $x$ for which $\sin x+i \cos 2 x$ and $\cos x-i \sin 2 x$ are conjugate to each other are
The values of $x$ for which $\sin x+i \cos 2 x$ and $\cos x-i \sin 2 x$ are conjugate to each other are
$x=n \pi \pm \frac{\pi}{6}$
None
$x=n \pi \pm \frac{\pi}{3}$
$x=\left(n+\frac{1}{2}\right) \pi$
Solution
Conjugate of $\cos x-i \sin 2 x$ is $\cos x+i \sin 2 x$
So, $\sin x+i \cos 2 x=\cos x+i \sin 2 x$
Comparing real and imaginary parts,
$\sin x=\cos x$ or $\tan x=1$
and $\cos 2 x=\sin 2 x$ or $\tan 2 x=1$
But for same value of $x$, both $\tan x$ and $\tan 2 x$ can not.
Hence, no solution is possible.