`sec^-1(sinx)` is real if

Options

A. ` x in (-oo, oo) `
B. ` x in [-1,1]`
C. ` x = (2n +1)(pi)/(2), n in I `
D. ` x= n pi, n in Z`

Answer

To determine when \( \sec^{-1}(\sin x) \) is real, we need to analyze the ranges and domains of the involved functions.

### Step-by-Step Solution:

1. **Understanding the Functions**:
- The function \( \sin x \) has a range of \( [-1, 1] \).
- The function \( \sec^{-1}(t) \) is defined for \( t \) such that \( t \leq -1 \) or \( t \geq 1 \).

2. **Finding Valid Inputs**:
- For \( \sec^{-1}(\sin x) \) to be real, \( \sin x \) must take values that are either \( \leq -1 \) or \( \geq 1 \).
- However, since the range of \( \sin x \) is limited to \( [-1, 1] \), the only valid values for \( \sin x \) that satisfy the domain of \( \sec^{-1} \) are \( -1 \) and \( 1 \).

3. **Solving for \( x \)**:
- We need to find the values of \( x \) for which \( \sin x = 1 \) and \( \sin x = -1 \).
- The equation \( \sin x = 1 \) occurs at:
\[ x = \frac{\pi}{2} + 2n\pi \quad (n \in \mathbb{Z}) \]
- The equation \( \sin x = -1 \) occurs at:
\[ x = \frac{3\pi}{2} + 2n\pi \quad (n \in \mathbb{Z}) \]

4. **Combining the Results**:
- The values of \( x \) where \( \sec^{-1}(\sin x) \) is real are:
\[ x = \frac{\pi}{2} + 2n\pi \quad \text{and} \quad x = \frac{3\pi}{2} + 2n\pi \quad (n \in \mathbb{Z}) \]
- These can be expressed as:
\[ x = (2n + 1)\frac{\pi}{2} \quad (n \in \mathbb{Z}) \]

5. **Conclusion**:
- Thus, \( \sec^{-1}(\sin x) \) is real if \( x \) is of the form \( (2n + 1)\frac{\pi}{2} \) for any integer \( n \).

### Final Answer:
The correct option is: \( x = (2n + 1)\frac{\pi}{2} \) for every \( n \in \mathbb{Z} \).