The domain of the function
`f(x)=sqrt( log_(2) sinx )` , is

Options

A. `{(2n+1)(pi)/(2): n in Z}`
B. `{(4n+1)pi)/(2): n in Z}`
C. `{(3n-1)(pi)/(2):n in Z }`
D. `{(npi)/(2) : n in Z}`

Explanation

To find the domain of the function \( f(x) = \sqrt{\log_2(\sin x)} \), we need to ensure that the expression inside the square root is non-negative. This means we must have:

\[ \log_2(\sin x) \geq 0 \]

### Step 1: Solve the logarithmic inequality

The inequality \( \log_2(\sin x) \geq 0 \) implies:

\[ \sin x \geq 2^0 = 1 \]

### Step 2: Analyze the sine function

The sine function, \( \sin x \), has a range of \([-1, 1]\). Therefore, the only way for \( \sin x \) to be greater than or equal to 1 is if:

\[ \sin x = 1 \]

### Step 3: Find the values of \( x \)

The sine function equals 1 at specific points. The general solution for \( \sin x = 1 \) occurs at:

\[ x = \frac{\pi}{2} + 2n\pi \]

where \( n \) is any integer.

### Conclusion: Domain of the function

Thus, the domain of the function \( f(x) = \sqrt{\log_2(\sin x)} \) is given by:

\[ x = \frac{\pi}{2} + 2n\pi, \quad n \in \mathbb{Z} \]

This means that the function is defined only at these discrete points.

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