The domain of the function `f(x) = (1)/(sqrt(4 + 3 sin x))` is :

Options

A. R
B. `[-(pi)/(2), (pi)/(2)]`
C. `R - {2 n pi, n in I}`
D. `(0, (pi)/(2))`

Answer

To find the domain of the function \( f(x) = \frac{1}{\sqrt{4 + 3 \sin x}} \), we need to ensure that the expression inside the square root is positive because the square root function is only defined for non-negative numbers, and we cannot have a zero in the denominator.

### Step-by-Step Solution:

1. **Identify the condition for the square root**:
We need the expression \( 4 + 3 \sin x \) to be greater than 0.
\[ 4 + 3 \sin x > 0 \]

2. **Rearranging the inequality**:
Subtract 4 from both sides:
\[ 3 \sin x > -4 \]

3. **Dividing by 3**:
Since 3 is positive, the direction of the inequality remains the same:
\[ \sin x > -\frac{4}{3} \]

4. **Analyzing the sine function**:
The sine function \( \sin x \) has a range of \([-1, 1]\). Therefore, the inequality \( \sin x > -\frac{4}{3} \) is always satisfied since \(-\frac{4}{3}\) is less than the minimum value of \(\sin x\).

5. **Finding the upper bound**:
Now we need to ensure that \( 4 + 3 \sin x \) does not equal zero:
\[ 4 + 3 \sin x \neq 0 \]
This leads to:
\[ 3 \sin x \neq -4 \quad \Rightarrow \quad \sin x \neq -\frac{4}{3} \]
Since \(-\frac{4}{3}\) is outside the range of \(\sin x\), this condition is automatically satisfied.

6. **Conclusion about the domain**:
Since \( 4 + 3 \sin x > 0 \) is always true for all \( x \), the function \( f(x) \) is defined for all real numbers.

Thus, the domain of the function \( f(x) = \frac{1}{\sqrt{4 + 3 \sin x}} \) is:
\[ \text{Domain} = \mathbb{R} \quad \text{(all real numbers)} \]