Find the domain of the following functions: `f(x)=sqrt(sin(cosx))`

Explanation

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

### Step-by-Step Solution:

1. **Identify the condition for the square root**:
The expression inside the square root, \( \sin(\cos x) \), must be greater than or equal to zero:
\[ \sin(\cos x) \geq 0 \]

2. **Set \( \cos x = \theta \)**:
Let \( \theta = \cos x \). The condition now becomes:
\[ \sin(\theta) \geq 0 \]

3. **Determine the intervals where \( \sin(\theta) \geq 0 \)**:
The sine function is non-negative in the intervals:
\[ \theta \in [0, \pi] + 2n\pi \quad (n \in \mathbb{Z}) \]
This means that \( \theta \) can take values in the ranges:
\[ [0, \pi], [2\pi, 3\pi], [4\pi, 5\pi], \ldots \]

4. **Relate \( \theta \) back to \( x \)**:
Since \( \theta = \cos x \), we need to find the values of \( x \) such that:
\[ \cos x \in [0, \pi] \]

5. **Find the corresponding \( x \) values**:
The cosine function is non-negative in the intervals:
\[ x \in [-\frac{\pi}{2} + 2n\pi, \frac{\pi}{2} + 2n\pi] \quad (n \in \mathbb{Z}) \]

6. **Conclusion**:
Therefore, the domain of the function \( f(x) = \sqrt{\sin(\cos x)} \) is:
\[ x \in \left[-\frac{\pi}{2} + 2n\pi, \frac{\pi}{2} + 2n\pi\right], \quad n \in \mathbb{Z} \]