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-by-Step Solution:

1. **Identify the Denominator**:
The denominator of the function is \( |\sin x| + \sin x \).

2. **Set the Denominator Not Equal to Zero**:
We need to find when:
\[ |\sin x| + \sin x \neq 0 \]

3. **Consider Cases for \( \sin x \)**:
We will analyze two cases based on the sign of \( \sin x \).

- **Case 1**: \( \sin x \geq 0 \)
- In this case, \( |\sin x| = \sin x \).
- Therefore, the denominator 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 any integer. Thus, \( \sin x \) is positive in the intervals:
\[ n\pi < x < (n+1)\pi \quad \text{for } n \text{ even} \]

- **Case 2**: \( \sin x < 0 \)
- Here, \( |\sin x| = -\sin x \).
- Thus, the denominator becomes:
\[ -\sin x + \sin x = 0 \]
- This case leads to a denominator of zero, which is not acceptable.

4. **Conclusion on the Domain**:
From Case 1, we conclude that the function is defined when \( \sin x \) is positive, which occurs in the intervals:
\[ (2n\pi, (2n+1)\pi) \quad \text{for } n \in \mathbb{Z} \]
Therefore, the domain of \( f(x) \) is:
\[ x \in \bigcup_{n \in \mathbb{Z}} (2n\pi, (2n+1)\pi) \]

### Final Answer:
The domain of \( f(x) = \frac{1}{|\sin x| + \sin x} \) is:
\[ \{ x \mid x \in (2n\pi, (2n+1)\pi), n \in \mathbb{Z} \} \]