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

Options

A. x=nx
B. `x=(n+(1)/(2))(pi)/(2)`
C. x=0
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 can follow these steps:

### Step 1: Define the complex numbers
Let:
\[ z_1 = \sin x + i \cos 2x \]
\[ z_2 = \cos x - i \sin 2x \]

### Step 2: Write the conjugate of \( z_1 \)
The conjugate of \( z_1 \) is:
\[ \overline{z_1} = \sin x - i \cos 2x \]

### Step 3: Set the conjugate equal to \( z_2 \)
For \( z_1 \) and \( z_2 \) to be conjugates, we need:
\[ \overline{z_1} = z_2 \]
This gives us the equation:
\[ \sin x - i \cos 2x = \cos x - i \sin 2x \]

### Step 4: Separate real and imaginary parts
From the equation above, we can separate the real and imaginary parts:
1. Real part:
\[ \sin x = \cos x \]
2. Imaginary part:
\[ -\cos 2x = -\sin 2x \quad \text{or} \quad \cos 2x = \sin 2x \]

### Step 5: Solve the real part equation
The equation \( \sin x = \cos x \) can be rewritten as:
\[ \tan x = 1 \]
This implies:
\[ x = n\pi + \frac{\pi}{4} \quad \text{for } n \in \mathbb{Z} \]

### Step 6: Solve the imaginary part equation
The equation \( \cos 2x = \sin 2x \) can be rewritten as:
\[ \tan 2x = 1 \]
This implies:
\[ 2x = m\pi + \frac{\pi}{4} \quad \text{for } m \in \mathbb{Z} \]
Thus:
\[ x = \frac{m\pi}{2} + \frac{\pi}{8} \]

### Step 7: Equate the two expressions for \( x \)
From the two equations we derived:
1. \( x = n\pi + \frac{\pi}{4} \)
2. \( x = \frac{m\pi}{2} + \frac{\pi}{8} \)

Setting them equal gives:
\[ n\pi + \frac{\pi}{4} = \frac{m\pi}{2} + \frac{\pi}{8} \]

### Step 8: Solve for \( n \) and \( m \)
Rearranging gives:
\[ n\pi - \frac{m\pi}{2} = \frac{\pi}{8} - \frac{\pi}{4} \]
\[ n\pi - \frac{m\pi}{2} = -\frac{\pi}{8} \]
Multiplying through by 8 to eliminate the fraction:
\[ 8n\pi - 4m\pi = -\pi \]
This simplifies to:
\[ 8n - 4m = -1 \]

### Step 9: Analyze the equation
This equation implies that \( 8n - 4m = -1 \) cannot hold for integer values of \( n \) and \( m \) since the left side is always even and the right side is odd. Therefore, there are no integer solutions.

### 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 conjugates.

### Final Answer
Hence, the answer is: **There exist no values of \( x \)**.

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