Find domain for `f(x)=sqrt(cos (sin x))`

Answer

To find the domain of the function \( f(x) = \sqrt{\cos(\sin x)} \), we need to ensure that the expression inside the square root is non-negative. This means we need to determine when \( \cos(\sin x) \geq 0 \).

### Step 1: Understand the condition for the square root
The square root function is defined only for non-negative values. Therefore, we require:
\[ \cos(\sin x) \geq 0 \]

### Step 2: Analyze the range of \( \sin x \)
The sine function, \( \sin x \), oscillates between -1 and 1 for all real numbers \( x \). Thus, we have:
\[ -1 \leq \sin x \leq 1 \]

### Step 3: Determine when \( \cos(y) \) is non-negative
The cosine function, \( \cos(y) \), is non-negative in the intervals:
\[ y \in [2k\pi, (2k + 1)\pi] \quad \text{for any integer } k \]
This means that \( \cos(y) \) is non-negative when \( y \) is in the first quadrant (from \( 0 \) to \( \frac{\pi}{2} \)) and in the fourth quadrant (from \( \frac{3\pi}{2} \) to \( 2\pi \)).

### Step 4: Find the values of \( \sin x \) that keep \( \cos(\sin x) \geq 0 \)
We need to find the values of \( \sin x \) that fall within the intervals where \( \cos(y) \) is non-negative:
1. For \( \sin x \) in the range \( [0, \frac{\pi}{2}] \):
- This corresponds to \( \sin x \) values from \( 0 \) to \( 1 \).
2. For \( \sin x \) in the range \( [\frac{3\pi}{2}, 2\pi] \):
- However, since \( \sin x \) only takes values from -1 to 1, we only consider \( \sin x = -1 \) to \( \sin x = 0 \).

### Step 5: Combine the intervals
Thus, the values of \( \sin x \) that keep \( \cos(\sin x) \geq 0 \) are:
- \( \sin x \) in the interval \( [-1, 0] \) and \( [0, 1] \).

### Conclusion: Identify the domain of \( f(x) \)
Since \( \sin x \) can take any value between -1 and 1 for all real numbers \( x \), the function \( f(x) = \sqrt{\cos(\sin x)} \) is defined for all real numbers \( x \).

Therefore, the domain of \( f(x) \) is:
\[ \text{Domain of } f(x) = \mathbb{R} \quad \text{(all real numbers)} \]