Domain of the function `f(x)=(1)/([sinx-1])` (where [.] denotes the greatest integer function) is

Options

A. `R~{2npi+(pi)/(2), n in l}`
B. `R~{npi+(pi)/(2), n in l)}`
C. `[2n pi, 2npi+(pi)/(2)]`
D. none of these

Answer

To find the domain of the function \( f(x) = \frac{1}{\lfloor \sin x - 1 \rfloor} \), we need to determine the values of \( x \) for which the function is defined.

### Step-by-Step Solution:

1. **Identify the function**:
The function is given as \( f(x) = \frac{1}{\lfloor \sin x - 1 \rfloor} \). The denominator \( \lfloor \sin x - 1 \rfloor \) must not be zero for \( f(x) \) to be defined.

2. **Analyze \( \sin x \)**:
The sine function, \( \sin x \), oscillates between -1 and 1. Therefore, \( \sin x - 1 \) will oscillate between -2 and 0:
\[ -2 \leq \sin x - 1 < 0 \]

3. **Determine when \( \lfloor \sin x - 1 \rfloor = 0 \)**:
The greatest integer function \( \lfloor y \rfloor \) is 0 when \( 0 \leq y < 1 \). Thus, we need to find when:
\[ 0 \leq \sin x - 1 < 1 \]
This simplifies to:
\[ 1 \leq \sin x < 2 \]
Since \( \sin x \) cannot exceed 1, the only point where \( \sin x - 1 = 0 \) is when \( \sin x = 1 \).

4. **Find the values of \( x \) where \( \sin x = 1 \)**:
The sine function equals 1 at:
\[ x = \frac{\pi}{2} + 2n\pi \quad (n \in \mathbb{Z}) \]
This means that \( f(x) \) is undefined at these points.

5. **Determine the domain of \( f(x) \)**:
The domain of \( f(x) \) is all real numbers except the points where \( \sin x = 1 \):
\[ \text{Domain of } f(x) = \mathbb{R} \setminus \left\{ \frac{\pi}{2} + 2n\pi \mid n \in \mathbb{Z} \right\} \]

### Final Answer:
The domain of the function \( f(x) = \frac{1}{\lfloor \sin x - 1 \rfloor} \) is:
\[ \mathbb{R} \setminus \left\{ \frac{\pi}{2} + 2n\pi \mid n \in \mathbb{Z} \right\} \]