To solve the problem, we need to analyze the function \( f(x) = \sin^{-1}(x) + |\sin^{-1}(x)| + \sin^{-1}(|x|) \) and determine the values of \( k \) for which the equation \( f(x) = k \) has two solutions.
### Step 1: Analyze the function \( f(x) \)
The function \( f(x) \) can be broken down based on the value of \( x \):
1. **For \( x \in [0, 1] \)**:
- Here, \( \sin^{-1}(x) \) is non-negative, so \( |\sin^{-1}(x)| = \sin^{-1}(x) \).
- Thus,
\[
f(x) = \sin^{-1}(x) + \sin^{-1}(x) + \sin^{-1}(x) = 3\sin^{-1}(x).
\]
2. **For \( x \in [-1, 0) \)**:
- Here, \( \sin^{-1}(x) \) is negative, so \( |\sin^{-1}(x)| = -\sin^{-1}(x) \).
- Thus,
\[
f(x) = \sin^{-1}(x) - \sin^{-1}(x) + \sin^{-1}(-x) = \sin^{-1}(-x) = -\sin^{-1}(x).
\]
### Step 2: Determine the range of \( f(x) \)
- For \( x \in [0, 1] \):
- As \( x \) increases from 0 to 1, \( \sin^{-1}(x) \) increases from 0 to \( \frac{\pi}{2} \).
- Therefore, \( f(x) \) increases from 0 to \( \frac{3\pi}{2} \).
- For \( x \in [-1, 0) \):
- As \( x \) decreases from 0 to -1, \( -\sin^{-1}(x) \) increases from 0 to \( \frac{\pi}{2} \).
- Therefore, \( f(x) \) increases from 0 to \( \frac{\pi}{2} \).
### Step 3: Graph the function
The graph of \( f(x) \) will have the following characteristics:
- It will be continuous and piecewise linear.
- It will have a maximum value of \( \frac{3\pi}{2} \) at \( x = 1 \) and a minimum value of 0 at \( x = 0 \).
### Step 4: Finding the values of \( k \)
For the equation \( f(x) = k \) to have two solutions, \( k \) must lie within the range of \( f(x) \) where the function is increasing in both intervals.
- From the analysis, we find that:
- For \( k \) in the interval \( (0, \frac{\pi}{2}) \), there will be two solutions: one in \( [0, 1] \) and one in \( [-1, 0) \).
- At \( k = 0 \), there is one solution \( x = 0 \).
- At \( k = \frac{\pi}{2} \), there is also one solution at \( x = -1 \).
### Conclusion
The true set of values of \( k \) for which the equation \( f(x) = k \) has two solutions is:
\[
k \in (0, \frac{\pi}{2}).
\]
