The domain of the function `f(x)=cos^(-1)[secx]`, where [x] denotes the greatest integer less than or equal to x, is

Options

A. `{x:x=(2n+1) pi, n in Z} cup { x : 2 m pi le x lt 2m pi+(pi)/(3), m in Z}`
B. `{x:x=2n pi , n in Z} cup { x: 2 m pi lt(x =2n pi) , n in Z} cup { x : 2 m pi lt x lt 2m pi +(pi)/(3), m in Z }`
C. `{x:(2n+1) pi, n in Z} cup { x : 2m pi lt x lt 2 m pi +(pi)/(3), m in Z}`
D. none of these

Answer

To find the domain of the function \( f(x) = \cos^{-1}[\sec x] \), where \([\cdot]\) denotes the greatest integer function, we need to determine the values of \(x\) for which the expression inside the cosine inverse is valid.

### Step-by-step solution:

1. **Understanding the range of \(\cos^{-1} x\)**:
The function \(\cos^{-1} x\) is defined for \(x\) in the interval \([-1, 1]\). Therefore, we need to find when \([\sec x]\) falls within this range.

2. **Finding the values of \([\sec x]\)**:
The greatest integer function \([\sec x]\) can take integer values. We need to find the integers \(n\) such that:
\[ -1 \leq [\sec x] \leq 1 \]
This means that \([\sec x]\) can be \(0\), \(1\), or \(-1\).

3. **Analyzing \([\sec x] = 0\)**:
For \([\sec x] = 0\), we have:
\[ 0 \leq \sec x < 1 \]
Since \(\sec x = \frac{1}{\cos x}\), this implies:
\[ \cos x > 1 \]
However, \(\cos x\) cannot be greater than \(1\), so there are no solutions in this case.

4. **Analyzing \([\sec x] = 1\)**:
For \([\sec x] = 1\), we have:
\[ 1 \leq \sec x < 2 \]
This translates to:
\[ 1 \leq \frac{1}{\cos x} < 2 \]
Taking the reciprocal gives:
\[ \frac{1}{2} < \cos x \leq 1 \]
The values of \(x\) for which \(\cos x\) lies in this range are:
\[ x \in [2n\pi, 2n\pi + \frac{\pi}{3}] \quad \text{for } n \in \mathbb{Z} \]

5. **Analyzing \([\sec x] = -1\)**:
For \([\sec x] = -1\), we have:
\[ -1 \leq \sec x < 0 \]
This translates to:
\[ -1 \leq \frac{1}{\cos x} < 0 \]
Taking the reciprocal gives:
\[ 0 < \cos x < -1 \]
However, \(\cos x\) cannot be negative, so there are no solutions in this case.

6. **Combining the results**:
The only valid range for \(x\) comes from the case where \([\sec x] = 1\):
\[ x \in [2n\pi, 2n\pi + \frac{\pi}{3}] \quad \text{for } n \in \mathbb{Z} \]

### Final Answer:
The domain of the function \( f(x) = \cos^{-1}[\sec x] \) is:
\[ \bigcup_{n \in \mathbb{Z}} [2n\pi, 2n\pi + \frac{\pi}{3}] \]