Question

If the complex numbers $sin x+icos ⁡ 2 x$ and $cos x-isin ⁡ 2 x$ are conjugate of each other, then the number of values of $x$ in the interval $[0,2 \pi)$ is equal to (where, $\left.i^{2}=-1\right)$

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (A)

$\sin x=\cos x$ and $\cos 2 x=\sin 2 x$
$\Rightarrow \sin x=\cos x \& 2 \cos ^{2} x-1=2 \sin x \cos x$
$\Rightarrow \sin x=\cos x \& 2 \cos ^{2} x-1=2 \cos ^{2} x$
$\Rightarrow \sin x=\cos x \&-1=0$ (Which is impossible)
Hence, no value of $x$ exists