The domain of the function ` f(x)= sqrt(cos x)` is

Options

A. `[ (3pi)/2 , 2pi]`
B. `[0, pi/2]`
C. `[ - pi/2 , pi/2]`
D. ` [ 0,pi/2] cup [ (3pi)/2 , 2pi]`

Explanation

To find the domain of the function \( f(x) = \sqrt{\cos x} \), we need to determine the values of \( x \) for which the expression under the square root, \( \cos x \), is non-negative (i.e., \( \cos x \geq 0 \)).

### Step-by-Step Solution:

1. **Identify the condition for the square root**:
The square root function is defined only for non-negative values. Therefore, we need:
\[ \cos x \geq 0 \]

2. **Determine where \( \cos x \) is non-negative**:
The cosine function is non-negative in the following intervals:
- From \( 0 \) to \( \frac{\pi}{2} \) (first quadrant)
- From \( \frac{3\pi}{2} \) to \( 2\pi \) (fourth quadrant)
- Additionally, cosine is also non-negative in the intervals:
- From \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \) (first quadrant)
- From \( -\frac{3\pi}{2} \) to \( -2\pi \) (fourth quadrant)

3. **Combine the intervals**:
The intervals where \( \cos x \) is non-negative can be combined:
\[ x \in [0, \frac{\pi}{2}] \cup [\frac{3\pi}{2}, 2\pi] \]
However, we can also express this in terms of periodicity since cosine is periodic with a period of \( 2\pi \).

4. **Write the domain in a general form**:
The general solution for the domain, considering the periodic nature of the cosine function, can be expressed as:
\[ x \in [2k\pi, (2k+1)\frac{\pi}{2}] \cup [(2k+1)\frac{3\pi}{2}, (2k+2)\pi] \quad \text{for } k \in \mathbb{Z} \]

5. **Conclusion**:
Thus, the domain of the function \( f(x) = \sqrt{\cos x} \) is:
\[ \text{Domain} = \{ x \mid \cos x \geq 0 \} \]