To solve the given problem step by step, we will first analyze the inequation and the equation provided, and then define the function accordingly.
### Step 1: Solve the Inequation
We need to solve the inequation:
\[
n(\sin x - \frac{1}{2})(\sin x - \frac{1}{\sqrt{2}}) < 0
\]
for \( x \) in the interval \([ \frac{\pi}{2}, \pi ]\).
#### Step 1.1: Find the roots of the equation
The roots of the equation \((\sin x - \frac{1}{2})(\sin x - \frac{1}{\sqrt{2}}) = 0\) are:
1. \(\sin x = \frac{1}{2} \Rightarrow x = \frac{\pi}{6}, \frac{5\pi}{6}\)
2. \(\sin x = \frac{1}{\sqrt{2}} \Rightarrow x = \frac{\pi}{4}\)
However, we are only interested in the interval \([ \frac{\pi}{2}, \pi ]\). Thus, the relevant roots are:
- \(\frac{5\pi}{6}\) (from \(\sin x = \frac{1}{2}\))
- \(\frac{\pi}{4}\) is not in the interval.
- \(\frac{\pi}{6}\) is not in the interval.
#### Step 1.2: Test intervals
We will test the sign of the expression in the intervals defined by the roots:
- Interval 1: \([ \frac{\pi}{2}, \frac{5\pi}{6} ]\)
- Interval 2: \([ \frac{5\pi}{6}, \pi ]\)
1. **For \( x = \frac{3\pi}{4} \) (in Interval 1)**:
\[
\sin \frac{3\pi}{4} = \frac{\sqrt{2}}{2} \Rightarrow \left(\frac{\sqrt{2}}{2} - \frac{1}{2}\right)\left(\frac{\sqrt{2}}{2} - \frac{1}{\sqrt{2}}\right) > 0
\]
2. **For \( x = \frac{11\pi}{12} \) (in Interval 2)**:
\[
\sin \frac{11\pi}{12} > \frac{1}{2} \text{ and } \sin \frac{11\pi}{12} > \frac{1}{\sqrt{2}} \Rightarrow \left(\sin \frac{11\pi}{12} - \frac{1}{2}\right)\left(\sin \frac{11\pi}{12} - \frac{1}{\sqrt{2}}\right) < 0
\]
Thus, the solution set \( A \) is:
\[
A = \left( \frac{5\pi}{6}, \pi \right)
\]
### Step 2: Solve the Equation
Now we solve the equation:
\[
\sin^{-1}(3x - 4x^3) = 3\sin^{-1}(x)
\]
#### Step 2.1: Analyze the equation
Using the property of inverse sine, we can rewrite:
\[
3x - 4x^3 = \sin(3\sin^{-1}(x))
\]
Using the triple angle formula for sine:
\[
\sin(3\theta) = 3\sin\theta - 4\sin^3\theta
\]
Thus, we have:
\[
3x - 4x^3 = 3x - 4x^3
\]
This equation holds for all \( x \) in the domain of \( \sin^{-1} \).
#### Step 2.2: Find the range of \( B \)
The domain of \( \sin^{-1}(x) \) is \([-1, 1]\). Therefore, we need to find the values of \( x \) such that:
\[
3x - 4x^3 \in [-1, 1]
\]
From the cubic equation, we can find the critical points and evaluate the function to determine the range.
After solving, we find:
\[
B = \left[-\frac{1}{2}, \frac{1}{2}\right]
\]
### Step 3: Define the Function \( f: A \to B \)
Now we define the function \( f \) such that:
- \( A = \left( \frac{5\pi}{6}, \pi \right) \)
- \( B = \left[-\frac{1}{2}, \frac{1}{2}\right] \)
### Conclusion:
The function \( f \) is defined from the interval \( A \) to the interval \( B \). The function is onto since the range of \( f \) covers all values from \( A \) to \( B \).
