Question

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

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

Answer 1

Answer: (D)

Given, numbers are conjugate to each other,
$\therefore \sin x +i \cos 2 x=\cos x-i \sin 2 x$
Equating real and imaginary parts, we get
$\sin x=\cos x$ and $\cos 2 x=\sin 2 x$
$\therefore \tan x=1$
$\Rightarrow x=\frac{\pi}{4},\, \frac{5 \pi}{4},\, \frac{9 \pi}{4}$ ...(i)
and $\tan 2x=1$
$\Rightarrow 2 x=\frac{\pi}{4},\, \frac{5 \pi}{4},\, \frac{9 \pi}{4}$ ...(ii)
$\Rightarrow x=\frac{\pi}{8}, \frac{5 \pi}{8}, \frac{9 \pi}{8},...$
There exists no value of $ x $ common in Eqs. (i) and (ii).