`sec^(-1)(sin x)` exist if

Options

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

Answer

To determine the values of \( x \) for which \( \sec^{-1}(\sin x) \) exists, 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 \( \sec^{-1}(y) \) is defined for \( y \) such that \( y \leq -1 \) or \( y \geq 1 \). This means that the values of \( y \) must either be less than or equal to -1 or greater than or equal to 1.

2. **Setting Up the Inequalities**:
- Since \( \sec^{-1}(\sin x) \) exists only when \( \sin x \) takes values in the domain of \( \sec^{-1} \), we need to find when:
\[ \sin x \leq -1 \quad \text{or} \quad \sin x \geq 1 \]

3. **Analyzing the Sine Function**:
- The sine function \( \sin x \) can never actually be less than -1 or greater than 1. Therefore, we check the boundary conditions:
- \( \sin x = 1 \) occurs at \( x = \frac{\pi}{2} + 2n\pi \) for \( n \in \mathbb{Z} \).
- \( \sin x = -1 \) occurs at \( x = \frac{3\pi}{2} + 2n\pi \) for \( n \in \mathbb{Z} \).

4. **Finding the Values of \( x \)**:
- The values of \( x \) for which \( \sec^{-1}(\sin x) \) exists are:
\[ x = \frac{\pi}{2} + 2n\pi \quad \text{(for } \sin x = 1\text{)} \]
\[ x = \frac{3\pi}{2} + 2n\pi \quad \text{(for } \sin x = -1\text{)} \]
- We can express these conditions in a more general form:
\[ x = (2n + 1)\frac{\pi}{2} \quad \text{for } n \in \mathbb{Z} \]

5. **Conclusion**:
- Therefore, the values of \( x \) for which \( \sec^{-1}(\sin x) \) exists are given by:
\[ x = (2n + 1)\frac{\pi}{2}, \quad n \in \mathbb{Z} \]