Question

The values of \(x\) for which \[ \sin x+i\cos 2x \] and \[ \cos x-i\sin 2x \] are conjugate to each other are

• A. \(x=n\pi \pm \frac{\pi}{6}\)
• B. None
• C. \(x=n\pi \pm \frac{\pi}{3}\)
• D. \(x=\left(n+\frac12\right)\pi\)

Hint:

If \[ a+ib \] and \[ c+id \] are conjugates, then \[ a=c \] and \[ b=-d. \] Always compare real and imaginary parts separately.

Answer 1

The Correct Option is B

Step 1: Write the given complex numbers.
Let \[ z_1=\sin x+i\cos 2x \] and \[ z_2=\cos x-i\sin 2x \] We are given that \(z_1\) and \(z_2\) are conjugates of each other.

Step 2: Use the condition for conjugate complex numbers.
If two complex numbers are conjugates, then: Real parts must be equal and Imaginary parts must be equal in magnitude and opposite in sign Thus, \[ \sin x=\cos x \] and \[ \cos 2x=\sin 2x \]

Step 3: Solve the first equation.
From \[ \sin x=\cos x \] we get \[ \tan x=1 \] Therefore, \[ x=n\pi+\frac{\pi}{4} \]

Step 4: Solve the second equation.
From \[ \cos 2x=\sin 2x \] we get \[ \tan 2x=1 \] Hence, \[ 2x=n\pi+\frac{\pi}{4} \] Therefore, \[ x=\frac{n\pi}{2}+\frac{\pi}{8} \]

Step 5: Compare both solutions.
From the first condition: \[ x=n\pi+\frac{\pi}{4} \] From the second condition: \[ x=\frac{n\pi}{2}+\frac{\pi}{8} \] There is no common value of \(x\) satisfying both conditions simultaneously.
Hence, no such \(x\) exists.

Step 6: Final conclusion.
Therefore, the correct answer is \[ \boxed{\text{None}} \]