The complex numbers `sin x - i cos 2x and cos x - i sin 2x` are conjugate to each other for

Options

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

Answer

To determine the values of \( x \) for which the complex numbers \( \sin x - i \cos 2x \) and \( \cos x - i \sin 2x \) are conjugate to each other, we start by setting the two complex numbers equal to each other.

### Step 1: Set the complex numbers equal
We have:
\[ \sin x - i \cos 2x = \cos x - i \sin 2x \]

### Step 2: Separate real and imaginary parts
From the equation above, we can separate the real and imaginary parts:
- Real part: \( \sin x = \cos x \)
- Imaginary part: \( -\cos 2x = -\sin 2x \) or \( \cos 2x = \sin 2x \)

### Step 3: Solve the first equation
The first equation is:
\[ \sin x = \cos x \]
This implies:
\[ \tan x = 1 \]
The general solutions for this equation are:
\[ x = n\pi + \frac{\pi}{4} \quad \text{for } n \in \mathbb{Z} \]

### Step 4: Solve the second equation
The second equation is:
\[ \cos 2x = \sin 2x \]
This implies:
\[ \tan 2x = 1 \]
The general solutions for this equation are:
\[ 2x = m\pi + \frac{\pi}{4} \quad \text{for } m \in \mathbb{Z} \]
Thus, solving for \( x \):
\[ x = \frac{m\pi}{2} + \frac{\pi}{8} \]

### Step 5: Find the intersection of the solutions
Now we need to find the intersection of the two sets of solutions:
1. From \( \sin x = \cos x \):
\[ x = n\pi + \frac{\pi}{4} \]
2. From \( \cos 2x = \sin 2x \):
\[ x = \frac{m\pi}{2} + \frac{\pi}{8} \]

### Step 6: Analyze the solutions
To find common values, we can equate the two expressions:
\[ n\pi + \frac{\pi}{4} = \frac{m\pi}{2} + \frac{\pi}{8} \]
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 7: Check for possible integer solutions
Now we check if there are integer values of \( n \) and \( m \) that satisfy this equation. After testing various integers, we find that there are no integer solutions that satisfy both conditions simultaneously.

### Conclusion
Thus, there are no values of \( x \) for which the complex numbers \( \sin x - i \cos 2x \) and \( \cos x - i \sin 2x \) are conjugate to each other.

### Final Answer
The answer is that there are no possible values for \( x \).