Question

The complex numbers \(\sin x+i \cos 2 x\) and \(\cos x\) - \(\mathrm{i} \sin 2 x,(\mathrm{i}=\sqrt{-1})\) are conjugate to each other for,

• A. \(x=\mathrm{n}\pi ,\mathrm{n}\in \mathbb{Z}\)
• B. \(x=(\mathrm{n}+\frac{1}{2})\pi ,\mathrm{n}\in \mathbb{Z}\)
• C. \(x=(3\mathrm{n}-1)\pi ,\mathrm{n}\in \mathbb{Z}\)
• D. No value of \(x\)

Hint:

Two complex numbers are conjugates when their real parts are equal and imaginary parts are opposite.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The conjugate of \(\sin x+i\cos2x\) is \(\sin x-i\cos2x\). Equate it to \(\cos x-i\sin2x\).

Step 2: Equate Parts
Real parts: \(\sin x=\cos x\). Imaginary parts: \(\cos2x=\sin2x\).

Step 3: Check Consistency
\(\sin x=\cos x\) gives \(x=n\pi+\dfrac{\pi}{4}\), so \(2x=2n\pi+\dfrac{\pi}{2}\).
Then \(\cos2x=0\) but \(\sin2x=1\), so the second equation fails.

Step 4: Result
No value of \(x\) satisfies both, option (D).

Final Answer:
There is no value of \(x\) for which they are conjugates. \[ \boxed{\text{(D) }\text{No value of }x} \]