Question

For what value of $x$, the complex numbers $\sin x+i\cos 2x$ and $\cos x-i\sin 2x$ are conjugate to each other.

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

Answer 1

The correct option is D

No values of $x$

Determining the value of$x$ by using the conjugate relation:

Two complex numbers will only be conjugate, if their imaginary parts differ by their sign.

Given, $\sin x+i\cos 2x$ is conjugate of $\cos x-i\sin 2x$.

That leads to. $\sin x-i\cos 2x=$ $\cos x-i\sin 2x$

Comparing real and imaginary parts separately,

$$\begin{gathered} \sin x=\cos x\dots \left(1\right) \\ -\cos 2x=-\sin 2x\dots \left(2\right) \end{gathered}$$

From equation $\left(1\right)$, we get,

$\tan x=1$

From equation $\left(2\right)$ we get,

$$\begin{gathered} \tan 2x=1 \\ \Rightarrow \frac{2\tan x}{1-{\tan }^{2}x}=1 \end{gathered}$$

Not satisfied by $\tan x=1$

Hence, no values of $x$ is possible.

Thus, the correct answer is option (D).