The domain of the function `f(x)=(1)/(sqrt(|cosx|+cosx))` is

Options

A. `[-2n pi, 2 n pi], n in Z`
B. `(2n pi,bar(2n+1) pi), n in Z`
C. `(((4n+1)pi)/(2),((4n+3)pi)/(2)), n in Z`
D. `(((4n-1)pi)/(2),((4n+1)pi)/(2)), n in Z`

Explanation

To find the domain of the function \( f(x) = \frac{1}{\sqrt{|\cos x| + \cos x}} \), we need to determine the values of \( x \) for which the expression under the square root is positive, as the square root function is only defined for non-negative values.

### Step-by-Step Solution:

1. **Identify the Expression Under the Square Root**:
We start with the expression \( |\cos x| + \cos x \).

2. **Analyze the Cases for \( \cos x \)**:
- **Case 1**: When \( \cos x \geq 0 \):
- Here, \( |\cos x| = \cos x \).
- Thus, \( |\cos x| + \cos x = \cos x + \cos x = 2\cos x \).
- **Case 2**: When \( \cos x < 0 \):
- Here, \( |\cos x| = -\cos x \).
- Thus, \( |\cos x| + \cos x = -\cos x + \cos x = 0 \).

3. **Determine When the Expression is Positive**:
- From **Case 1**, we need \( 2\cos x > 0 \), which simplifies to \( \cos x > 0 \).
- From **Case 2**, the expression equals zero, which does not contribute to the domain since we cannot take the square root of zero in the denominator.

4. **Find the Intervals Where \( \cos x > 0 \)**:
- The cosine function is positive in the first and fourth quadrants.
- This occurs in the intervals:
- \( 0 < x < \frac{\pi}{2} \) (first quadrant)
- \( \frac{3\pi}{2} < x < 2\pi \) (fourth quadrant)

5. **Generalize the Intervals**:
- The cosine function is periodic with a period of \( 2\pi \).
- Therefore, we can express the intervals where \( \cos x > 0 \) as:
- \( x \in (2n\pi, 2n\pi + \frac{\pi}{2}) \) and \( x \in (2n\pi + \frac{3\pi}{2}, 2(n+1)\pi) \) for any integer \( n \).

6. **Final Domain Representation**:
- The domain of the function \( f(x) \) is:
\[ x \in \bigcup_{n \in \mathbb{Z}} \left(2n\pi, 2n\pi + \frac{\pi}{2}\right) \cup \left(2n\pi + \frac{3\pi}{2}, 2(n+1)\pi\right) \]