Solve for real $x$: ${\tan }^{-1}(2x)={\sin }^{-1}(x)$. List all possible solutions.

Step-by-step solution

1. Let $\theta ={\sin }^{-1}x$. Then $x=\sin \theta$ and $\theta \in [-\frac{\pi }{2},\frac{\pi }{2}]$.
2. The given equation becomes ${\tan }^{-1}(2\sin \theta )=\theta$.
3. Take tangent on both sides: $2\sin \theta =\tan \theta =\frac{\sin \theta }{\cos \theta }$.
4. Multiply by $\cos \theta$: $2\sin \theta \cos \theta =\sin \theta$.
5. Factor $\sin \theta$: $\sin \theta (2\cos \theta -1)=0$.
6. Case 1: $\sin \theta =0\Rightarrow \theta =0\Rightarrow x=\sin 0=0$.
7. Case 2: $2\cos \theta -1=0\Rightarrow \cos \theta =\frac{1}{2}$.
8. In the principal interval, $\cos \theta =\frac{1}{2}$ gives $\theta =\pm \frac{\pi }{3}$.
9. Corresponding $x$ values: $x=\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}$ and $x=\sin \left(-\frac{\pi }{3}\right)=-\frac{\sqrt{3}}{2}$.
10. All three values satisfy the original equation because ${\tan }^{-1}(2x)$ lies in $(-\frac{\pi }{2},\frac{\pi }{2})$ for each $x$.

Concept tested: Solving equations containing inverse trig functions

Published 13 September 2026 · How this solution was prepared and checked