Solve for real $x$: ${\sin }^{-1}x+{\cos }^{-1}(2x)=\frac{\pi }{2}$. State all possible solutions and any restrictions on $x$.

Step-by-step solution

1. Recall the identity ${\cos }^{-1}t=\frac{\pi }{2}-{\sin }^{-1}t$ for $t\in [-1,1]$.
2. Apply it: ${\sin }^{-1}x+\left(\frac{\pi }{2}-{\sin }^{-1}(2x)\right)=\frac{\pi }{2}$.
3. Cancel $\frac{\pi }{2}$ from both sides, obtaining ${\sin }^{-1}x={\sin }^{-1}(2x)$.
4. Since ${\sin }^{-1}$ is one‑to‑one on $[-1,1]$, we must have $x=2x$.
5. Solve: $x=2x\Rightarrow x=0$.
6. Check domain: both $x$ and $2x$ must lie in $[-1,1]$, which is true for $x=0$.
7. Hence the only real solution is $x=0$.

Concept tested: Solving equation with inverse sine and cosine

Published 16 September 2026 · How this solution was prepared and checked