To solve the equation $2 \sin^{-1}(x) = \frac{\pi}{2}$, follow these steps:

1. Divide both sides by 2 to isolate $\sin^{-1}(x)$: $\sin^{-1}(x) = \frac{\pi}{4}$

2. Apply the sine function to both sides to eliminate the inverse sine: $x = \sin\left(\frac{\pi}{4}\right)$

3. Evaluate $\sin\left(\frac{\pi}{4}\right)$. We know: $\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$

Thus, the solution is: $x = \frac{\sqrt{2}}{2}$

Final Answer:

$x = \frac{\sqrt{2}}{2}$

Would you like further clarification or a more detailed explanation?

5 Related Questions:

1. What is the range of the inverse sine function?
2. How would the solution change if the equation were $3 \sin^{-1}(x) = \frac{\pi}{2}$?
3. How does the sine function behave in different quadrants of the unit circle?
4. Can you solve $\sin^{-1}(x) = \frac{\pi}{6}$?
5. How would you solve $2 \cos^{-1}(x) = \frac{\pi}{2}$?

Tip:

When solving equations involving inverse trigonometric functions, always check that the solution lies within the defined range of the inverse function.