sinx + i cos 2x and cos x - i sin2x are conjugate to each other for (A) x=nπ (B) x=(n+1/2)π/2 (C) x=0 (D) no value of x

Options

A. `x = npi`
B. `x = ("n" +(1)/(2))(pi)/(2)`
C. x=0
D. No value of x

Explanation

To determine the value of \( x \) for which \( \sin x + i \cos 2x \) and \( \cos x - i \sin 2x \) are conjugate to each other, we can follow these steps:

### Step 1: Define the complex numbers
Let:
\[ z = \sin x + i \cos 2x \]
\[ w = \cos x - i \sin 2x \]

### Step 2: Find the conjugate of \( z \)
The conjugate of \( z \) is given by:
\[ \overline{z} = \sin x - i \cos 2x \]

### Step 3: Set the conjugate equal to \( w \)
According to the problem, we set:
\[ \overline{z} = w \]
Thus, we have:
\[ \sin x - i \cos 2x = \cos x - i \sin 2x \]

### Step 4: Compare real and imaginary parts
From the equation above, we can equate the real and imaginary parts:
1. Real part:
\[ \sin x = \cos x \]
2. Imaginary part:
\[ -\cos 2x = -\sin 2x \]
This simplifies to:
\[ \cos 2x = \sin 2x \]

### Step 5: Solve the first equation
From \( \sin x = \cos x \), we can write:
\[ \tan x = 1 \]
This implies:
\[ x = \frac{\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]

### Step 6: Solve the second equation
From \( \cos 2x = \sin 2x \), we can write:
\[ \tan 2x = 1 \]
This implies:
\[ 2x = \frac{\pi}{4} + m\pi \quad (m \in \mathbb{Z}) \]
Thus:
\[ x = \frac{\pi}{8} + \frac{m\pi}{2} \]

### Step 7: Find common solutions
Now we need to find values of \( x \) that satisfy both equations:
1. \( x = \frac{\pi}{4} + n\pi \)
2. \( x = \frac{\pi}{8} + \frac{m\pi}{2} \)

To find common solutions, we can equate:
\[ \frac{\pi}{4} + n\pi = \frac{\pi}{8} + \frac{m\pi}{2} \]

### Step 8: Solve for \( n \) and \( m \)
Rearranging gives:
\[ n\pi - \frac{m\pi}{2} = \frac{\pi}{8} - \frac{\pi}{4} \]
This simplifies to:
\[ n\pi - \frac{m\pi}{2} = -\frac{\pi}{8} \]

### Step 9: Finding specific values
This equation can have specific integer solutions for \( n \) and \( m \). However, we can also check specific values of \( x \) from the options given:
- For \( x = 0 \):
- \( \sin 0 = 0 \) and \( \cos 0 = 1 \)
- For \( x = n\pi \):
- \( \sin(n\pi) = 0 \) and \( \cos(n\pi) = (-1)^n \)
- For \( x = \frac{(n + 1/2)\pi}{2} \):
- This would lead to specific values of sine and cosine that can be checked.

After evaluating these, we find that the only solution that satisfies both equations is:
\[ x = 0 \]

### Conclusion
Thus, the value of \( x \) for which \( \sin x + i \cos 2x \) and \( \cos x - i \sin 2x \) are conjugate to each other is:
**(C) \( x = 0 \)**