Solve sin⁻¹x + sin⁻¹2x = π/2. Find the domain, solve by isolating sin⁻¹2x and taking the sine, and…

Step 1. The domain. $|x| \le \dfrac{1}{2}$.

Step 2. Take the sine. $\sin^{-1}2x = \cos^{-1}x$ gives $2x = \sqrt{1 - x^2}$. Squaring: $4x^2 = 1 - x^2$, so $x^2 = \dfrac{1}{5}$ and $x = \pm\dfrac{1}{\sqrt{5}} = \pm 0.4472$. Both are inside $|x| \le \dfrac{1}{2}$.

Step 3. Test. For $x = 0.4472$: $\sin^{-1}0.4472 + \sin^{-1}0.8944 = 0.4636 + 1.1071 = 1.5708 = \dfrac{\pi}{2}$. It is a solution. For $x = -0.4472$ the left side is $-0.4636 - 1.1071 = -1.5708 = -\dfrac{\pi}{2}$. This is false, and it is the reason for the minus sign: $2x = \sqrt{1 - x^2}$ needs $x \ge 0$, and squaring lost that.