The complex numbers $\sin x+i\cos 2x$ and $\cos x-i\sin 2x$ are conjugate to each other, for
Step-by-Step Solution
Key Concept: Two complex numbers are conjugates iff their real parts are equal AND their imaginary parts are negatives of each other simultaneously. Here the two conditions lead to incompatible equations.
**Step 1: Write the conjugate condition**
If $\sin x+i\cos 2x$ and $\cos x-i\sin 2x$ are conjugates, then $\cos x-i\sin 2x=\overline{\sin x+i\cos 2x}=\sin x-i\cos 2x$.
**Step 2: Equate real parts**
$\cos x=\sin x \Rightarrow x=\dfrac{\pi}{4}+\dfrac{n\pi}{2}$.
**Step 3: Equate imaginary parts**
$-\sin 2x=-\cos 2x \Rightarrow \tan 2x=1 \Rightarrow x=\dfrac{\pi}{8}+\dfrac{m\pi}{2}$.
**Step 4: Check compatibility**
The two families $\dfrac{\pi}{4}+\dfrac{n\pi}{2}$ and $\dfrac{\pi}{8}+\dfrac{m\pi}{2}$ never coincide for any integers $n,m$. So there is no value of $x$.
Correct Answer: 4