Solve `log_(|sin x|) (1+cos x)=2`.

Explanation

To solve the equation \( \log_{|\sin x|}(1 + \cos x) = 2 \), we will follow these steps:

### Step 1: Rewrite the logarithmic equation in exponential form
Using the property of logarithms, we can rewrite the equation as:
\[ 1 + \cos x = |\sin x|^2 \]
This is because if \( \log_b(a) = c \), then \( a = b^c \).

### Step 2: Simplify the equation
Recall that \( |\sin x|^2 = \sin^2 x \). Therefore, we can rewrite the equation as:
\[ 1 + \cos x = \sin^2 x \]

### Step 3: Use the Pythagorean identity
We know that \( \sin^2 x + \cos^2 x = 1 \). We can express \( \sin^2 x \) in terms of \( \cos x \):
\[ \sin^2 x = 1 - \cos^2 x \]
Substituting this into our equation gives:
\[ 1 + \cos x = 1 - \cos^2 x \]

### Step 4: Rearrange the equation
Now, we can rearrange the equation:
\[ \cos^2 x + \cos x = 0 \]

### Step 5: Factor the equation
Factoring out \( \cos x \) gives:
\[ \cos x (\cos x + 1) = 0 \]

### Step 6: Solve for \( \cos x \)
Setting each factor to zero gives us:
1. \( \cos x = 0 \)
2. \( \cos x + 1 = 0 \) which simplifies to \( \cos x = -1 \)

### Step 7: Find the values of \( x \)
1. For \( \cos x = 0 \):
- This occurs at \( x = \frac{\pi}{2} + n\pi \) where \( n \) is any integer.

2. For \( \cos x = -1 \):
- This occurs at \( x = \pi + 2n\pi \) where \( n \) is any integer.

### Step 8: Check the validity of solutions
We need to ensure that the base of the logarithm \( |\sin x| \) is valid (i.e., \( |\sin x| > 0 \)):
- For \( x = \frac{\pi}{2} + n\pi \), \( \sin x = (-1)^n \) which is either \( 1 \) or \( -1 \), hence \( |\sin x| = 1 \) (valid).
- For \( x = \pi + 2n\pi \), \( \sin x = 0 \), hence \( |\sin x| = 0 \) (not valid).

### Conclusion
The only valid solutions are:
\[ x = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]