Find the number of solution of `log_(1/2) |sinx| = 2 - log_(1/2) |cosx|, x in [0, 2pi]`

Options

A. 2
B. 4
C. 6
D. 8

Explanation

To solve the equation \( \log_{1/2} |\sin x| = 2 - \log_{1/2} |\cos x| \) for \( x \) in the interval \([0, 2\pi]\), we can follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ \log_{1/2} |\sin x| = 2 - \log_{1/2} |\cos x| \]
We can rearrange this to bring the logarithm terms to one side:
\[ \log_{1/2} |\sin x| + \log_{1/2} |\cos x| = 2 \]

### Step 2: Combine the logarithms
Using the property of logarithms that states \( \log_b a + \log_b c = \log_b (a \cdot c) \), we can combine the left-hand side:
\[ \log_{1/2} (|\sin x| \cdot |\cos x|) = 2 \]

### Step 3: Exponentiate both sides
To eliminate the logarithm, we exponentiate both sides with base \( \frac{1}{2} \):
\[ |\sin x| \cdot |\cos x| = \left(\frac{1}{2}\right)^2 \]
This simplifies to:
\[ |\sin x| \cdot |\cos x| = \frac{1}{4} \]

### Step 4: Use the double angle identity
We know that \( \sin(2x) = 2 \sin x \cos x \). Thus, we can express our equation in terms of \( \sin(2x) \):
\[ \frac{1}{2} |\sin(2x)| = \frac{1}{4} \]
Multiplying both sides by 2 gives:
\[ |\sin(2x)| = \frac{1}{2} \]

### Step 5: Solve for \( 2x \)
The equation \( |\sin(2x)| = \frac{1}{2} \) implies two cases:
1. \( \sin(2x) = \frac{1}{2} \)
2. \( \sin(2x) = -\frac{1}{2} \)

### Step 6: Find solutions for \( 2x \)
**Case 1:** \( \sin(2x) = \frac{1}{2} \)
The general solutions for this are:
\[ 2x = \frac{\pi}{6} + 2k\pi \quad \text{and} \quad 2x = \frac{5\pi}{6} + 2k\pi \]
Dividing by 2 gives:
\[ x = \frac{\pi}{12} + k\pi \quad \text{and} \quad x = \frac{5\pi}{12} + k\pi \]

**Case 2:** \( \sin(2x) = -\frac{1}{2} \)
The general solutions for this are:
\[ 2x = \frac{7\pi}{6} + 2k\pi \quad \text{and} \quad 2x = \frac{11\pi}{6} + 2k\pi \]
Dividing by 2 gives:
\[ x = \frac{7\pi}{12} + k\pi \quad \text{and} \quad x = \frac{11\pi}{12} + k\pi \]

### Step 7: Determine valid solutions in \([0, 2\pi]\)
Now we need to find the values of \( x \) in the interval \([0, 2\pi]\):

- From \( x = \frac{\pi}{12} + k\pi \):
- For \( k = 0 \): \( x = \frac{\pi}{12} \)
- For \( k = 1 \): \( x = \frac{\pi}{12} + \pi = \frac{13\pi}{12} \)

- From \( x = \frac{5\pi}{12} + k\pi \):
- For \( k = 0 \): \( x = \frac{5\pi}{12} \)
- For \( k = 1 \): \( x = \frac{5\pi}{12} + \pi = \frac{17\pi}{12} \)

- From \( x = \frac{7\pi}{12} + k\pi \):
- For \( k = 0 \): \( x = \frac{7\pi}{12} \)
- For \( k = 1 \): \( x = \frac{7\pi}{12} + \pi = \frac{19\pi}{12} \)

- From \( x = \frac{11\pi}{12} + k\pi \):
- For \( k = 0 \): \( x = \frac{11\pi}{12} \)
- For \( k = 1 \): \( x = \frac{11\pi}{12} + \pi = \frac{23\pi}{12} \)

### Step 8: Count the valid solutions
Now we list all the valid solutions:
1. \( \frac{\pi}{12} \)
2. \( \frac{5\pi}{12} \)
3. \( \frac{7\pi}{12} \)
4. \( \frac{11\pi}{12} \)
5. \( \frac{13\pi}{12} \)
6. \( \frac{17\pi}{12} \)
7. \( \frac{19\pi}{12} \)
8. \( \frac{23\pi}{12} \)

Thus, there are a total of **8 solutions** in the interval \([0, 2\pi]\).

### Final Answer
The number of solutions is \( \boxed{8} \).