the complex number sinx+icos2x and cosx+isin2x are conjugate to each other for what values of x

since the conjugate of $z=a+ib is \overline{z} =a-ib$
conjugate of $\sin x+i\cos 2x is \sin x-i\cos 2x$
therefore
$$\begin{gathered} \sin x-i\cos 2x=\cos x+i\sin 2x \\ (\sin x-\cos x)-i(\sin 2x+\cos 2x)=0 \end{gathered}$$
equating the real and imaginary both part to zero:
$$\begin{gathered} \sin x-\cos x=0 \& \sin 2x+\cos 2x=0 \\ \sin x=\cos x \& \sin 2x=-\cos 2x \\ \frac{\sin x}{\cos x}=1 \& \frac{\sin 2x}{\cos 2x}=-1 \\ \tan x=1 \& \tan 2x=-1 \\ x=n\mathrm{\pi }+\frac{\mathrm{\pi }}{4} \& 2\mathrm{x}=\mathrm{n\pi }-\frac{\mathrm{\pi }}{4} \\ \mathrm{x}=\mathrm{n\pi }+\frac{\mathrm{\pi }}{4} \& \mathrm{x}=\frac{\mathrm{n\pi }}{2}-\frac{\mathrm{\pi }}{8} \end{gathered}$$
thus we will not get any value of x, which satisfy both the equation simultaneously.
thus no such value of x exist.
hope this helps you