If the complex numbers `sinx+icos 2x` and `cosx-isin2x` are conjugate of each other, then the number of values of x in the inverval `[0, 2pi)` is equal to (where, `i^(2)=-1`)

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the problem, we need to determine the values of \( x \) in the interval \([0, 2\pi)\) for which the complex numbers \( \sin x + i \cos 2x \) and \( \cos x - i \sin 2x \) are conjugates of each other.

### Step 1: Set up the equation for conjugates
For two complex numbers \( z_1 = a + ib \) and \( z_2 = c + id \) to be conjugates, we must have:
- The real part of \( z_1 \) equal to the real part of \( z_2 \)
- The imaginary part of \( z_1 \) equal to the negative of the imaginary part of \( z_2 \)

In our case:
- \( z_1 = \sin x + i \cos 2x \)
- \( z_2 = \cos x - i \sin 2x \)

From this, we can equate:
1. Real parts: \( \sin x = \cos x \)
2. Imaginary parts: \( \cos 2x = -(-\sin 2x) \) which simplifies to \( \cos 2x = \sin 2x \)

### Step 2: Solve the real part equation
From \( \sin x = \cos x \), we can write:
\[ \tan x = 1 \]
The general solution for \( \tan x = 1 \) is:
\[ x = \frac{\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]
In the interval \([0, 2\pi)\), the values of \( x \) are:
- \( x = \frac{\pi}{4} \)
- \( x = \frac{5\pi}{4} \)

### Step 3: Solve the imaginary part equation
From \( \cos 2x = \sin 2x \), we can write:
\[ \tan 2x = 1 \]
The general solution for \( \tan 2x = 1 \) is:
\[ 2x = \frac{\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]
Thus:
\[ x = \frac{\pi}{8} + \frac{n\pi}{2} \]
In the interval \([0, 2\pi)\), we can find the values of \( x \):
- For \( n = 0 \): \( x = \frac{\pi}{8} \)
- For \( n = 1 \): \( x = \frac{5\pi}{8} \)
- For \( n = 2 \): \( x = \frac{9\pi}{8} \)
- For \( n = 3 \): \( x = \frac{13\pi}{8} \)

### Step 4: Combine the solutions
Now we have the solutions from both parts:
- From \( \sin x = \cos x \): \( x = \frac{\pi}{4}, \frac{5\pi}{4} \)
- From \( \tan 2x = 1 \): \( x = \frac{\pi}{8}, \frac{5\pi}{8}, \frac{9\pi}{8}, \frac{13\pi}{8} \)

### Step 5: Count the unique solutions
The unique solutions in the interval \([0, 2\pi)\) are:
- \( \frac{\pi}{4} \)
- \( \frac{5\pi}{4} \)
- \( \frac{\pi}{8} \)
- \( \frac{5\pi}{8} \)
- \( \frac{9\pi}{8} \)
- \( \frac{13\pi}{8} \)

Thus, the total number of unique values of \( x \) is **6**.

### Final Answer
The number of values of \( x \) in the interval \([0, 2\pi)\) is **6**.