Consider the function `f(x) = sin^(-1)x`, having principal value branch `[(pi)/(2), (3pi)/(2)] and g(x) = cos^(-1)x`, having principal value brach `[0, pi]`
For `sin^(-1) x lt (3pi)/(4)`, solution set of x is

Options

A. `((1)/(sqrt2), 1]`
B. `(-(1)/(sqrt2), -1]`
C. `[-(1)/(sqrt2), (1)/(sqrt2)]`
D. `[-1, 1]`

Explanation

To solve the problem, we need to find the solution set of \( x \) for the inequality \( \sin^{-1} x < \frac{3\pi}{4} \).

### Step-by-step Solution:

1. **Understanding the Inequality**:
We start with the inequality:
\[ \sin^{-1} x < \frac{3\pi}{4} \]

2. **Finding the Corresponding Sine Values**:
To eliminate the inverse sine function, we take the sine of both sides. However, we need to ensure that we are within the valid range of the sine function. Since \( \sin^{-1} x \) is defined for \( x \in [-1, 1] \), we can take the sine of both sides:
\[ x < \sin\left(\frac{3\pi}{4}\right) \]

3. **Calculating \( \sin\left(\frac{3\pi}{4}\right) \)**:
We know that:
\[ \sin\left(\frac{3\pi}{4}\right) = \sin\left(\pi - \frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]

4. **Setting the Upper Bound**:
Thus, we have:
\[ x < \frac{1}{\sqrt{2}} \]

5. **Finding the Lower Bound**:
Since \( \sin^{-1} x \) is defined for \( x \in [-1, 1] \), we also have the lower limit:
\[ -1 \leq x \leq 1 \]

6. **Combining the Inequalities**:
Therefore, combining the results, we have:
\[ -1 \leq x < \frac{1}{\sqrt{2}} \]

### Final Solution Set:
The solution set of \( x \) is:
\[ x \in [-1, \frac{1}{\sqrt{2}}) \]