The domain of `f(x)=(1)/(|sin x|+sin x|)" is:`

Options

A. R
B. `underset(n in Z) uu ((2n+1)pi, (2n+2) pi)`
C. `underset(n in Z) uu (2npi, (2n+1)pi)`
D. `phi`

Answer

To find the domain of the function \( f(x) = \frac{1}{|\sin x| + \sin x} \), we need to determine when the denominator is not equal to zero, as division by zero is undefined.

### Step 1: Set the Denominator Not Equal to Zero
We need to solve the inequality:
\[ |\sin x| + \sin x \neq 0 \]

### Step 2: Analyze the Cases for \( \sin x \)
We will consider two cases based on the value of \( \sin x \).

**Case 1: \( \sin x \geq 0 \)**
- In this case, \( |\sin x| = \sin x \).
- Therefore, the expression becomes:
\[ \sin x + \sin x = 2\sin x \]
- For this to be non-zero:
\[ 2\sin x \neq 0 \implies \sin x \neq 0 \]
- The sine function is zero at \( x = n\pi \) where \( n \) is an integer. Thus, \( \sin x \) is non-zero when \( x \) is not of the form \( n\pi \).

**Case 2: \( \sin x < 0 \)**
- Here, \( |\sin x| = -\sin x \).
- The expression becomes:
\[ -\sin x + \sin x = 0 \]
- This means that when \( \sin x < 0 \), the denominator is equal to zero, which is not allowed.

### Step 3: Combine the Results
From the analysis:
- In Case 1, \( \sin x \) must not be zero, which excludes \( x = n\pi \).
- In Case 2, \( \sin x < 0 \) leads to a denominator of zero, so we exclude all values where \( \sin x \) is negative.

### Step 4: Determine the Valid Intervals
The sine function is positive in the intervals:
\[ (2n\pi, (2n + 1)\pi) \quad \text{for } n \in \mathbb{Z} \]
Thus, the domain of \( f(x) \) is:
\[ x \in \bigcup_{n \in \mathbb{Z}} (2n\pi, (2n + 1)\pi) \]

### Conclusion
The domain of \( f(x) = \frac{1}{|\sin x| + \sin x} \) is:
\[ \{ x \in \mathbb{R} : x \neq n\pi \text{ and } x \in (2n\pi, (2n + 1)\pi) \text{ for } n \in \mathbb{Z} \} \]