Find the domain of `f(x)=sqrt(sinx)+sqrt(16-x^2)`

Explanation

To find the domain of the function \( f(x) = \sqrt{\sin x} + \sqrt{16 - x^2} \), we need to ensure that both square root expressions are defined and non-negative.

### Step 1: Analyze the first term \( \sqrt{\sin x} \)

The expression \( \sqrt{\sin x} \) is defined when \( \sin x \geq 0 \). The sine function is non-negative in the intervals:
- \( [0, \pi] \)
- \( [2\pi, 3\pi] \)
- and so on.

In general, \( \sin x \) is non-negative in the intervals:
- \( [2n\pi, (2n+1)\pi] \) for any integer \( n \).

### Step 2: Analyze the second term \( \sqrt{16 - x^2} \)

The expression \( \sqrt{16 - x^2} \) is defined when \( 16 - x^2 \geq 0 \), which simplifies to:
\[ x^2 \leq 16 \]
This means:
\[ -4 \leq x \leq 4 \]

### Step 3: Combine the conditions

Now we need to find the intersection of the intervals from Step 1 and Step 2.

1. From the second term, we have \( x \) must be in the interval \( [-4, 4] \).
2. From the first term, we need to consider the intervals where \( \sin x \geq 0 \) within \( [-4, 4] \).

### Step 4: Identify specific intervals where \( \sin x \geq 0 \) within \( [-4, 4] \)

1. The sine function is non-negative in the interval \( [0, \pi] \) which approximately equals \( [0, 3.14] \).
2. The sine function is also non-negative in the interval \( [-\pi, 0] \) which approximately equals \( [-3.14, 0] \).

### Step 5: Determine the final domain

Now we combine these intervals:
- From \( [-4, -3.14] \) (where \( \sin x \geq 0 \))
- From \( [0, 3.14] \) (where \( \sin x \geq 0 \))

Thus, the domain of \( f(x) \) is:
\[ x \in [-\pi, 0] \cup [0, \pi] \]

### Final Answer
The domain of \( f(x) = \sqrt{\sin x} + \sqrt{16 - x^2} \) is:
\[ [-\pi, 0] \cup [0, \pi] \]