Question

$\sin x+i\cos 2x$ and $\cos x-i\sin 2x$ are conjugate to each other for:

• A. $x=0$
• B. $x=n\pi$
• C. $x=(n+\frac{1}{2})\frac{\pi }{2}$
• D. No value of $x$

Answer 1

The correct option is D No value of $x$

Given : $\sin x+i\cos 2x$ and

$\cos x-i\sin 2x$ are conjugate.

Conjugate of ($\sin x+i\cos 2x$)

$=\sin x-i\cos 2x$

Now, comparing both of the

Conjugates, we get

$\sin x-i\cos 2x=\cos x-i\sin 2x$

Comparing the real and imaginary parts, we get

$\sin x=\cos x$ and $\cos 2x=\sin 2x$

$\Rightarrow \tan x=1$ and $\tan 2x=1$

$\Rightarrow \tan x=1$ and $\tan 2x=1$

Now, $\tan 2x=1$

$\Rightarrow \frac{2\tan x}{1-{\tan }^{2}x}=1$, which is not satisfied

by $\tan x=1$

Hence, no value of $x$ is possible

Hence, Option $\left(D\right)$ is correct.