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,
$\therefore \sin x+i\cos 2x=\cos x-i\sin 2x$
Equating real and imaginary parts, we get
$\sin x=\cos x$ and $\cos 2x=\sin 2x$
$\therefore \tan x=1$
$\Rightarrow x=\frac{\pi }{4},\frac{5\pi }{4},\frac{9\pi }{4}$ ...(i)
and $\tan 2x=1$
$\Rightarrow 2x=\frac{\pi }{4},\frac{5\pi }{4},\frac{9\pi }{4}$ ...(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).