Solve for real $x$: ${\tan }^{-1}x-{\sin }^{-1}\frac{2x}{1+x^2}=0$. List all possible values of $x$.

Step-by-step solution

1. Let $\theta ={\tan }^{-1}x$, so $x=\tan \theta$ and $\theta \in (-\pi /2,\pi /2)$.
2. Observe that $\frac{2x}{1+x^2}=\frac{2\tan \theta }{1+{\tan }^2\theta }=\sin 2\theta$.
3. Hence the equation becomes $\theta ={\sin }^{-1}(\sin 2\theta )$.
4. Because ${\sin }^{-1}$ returns a value in $[-\pi /2,\pi /2]$, we consider the piecewise definition:
5. If $2\theta \in [-\pi /2,\pi /2]$, then ${\sin }^{-1}(\sin 2\theta )=2\theta$, giving $\theta =2\theta \Rightarrow \theta =0$.
6. If $2\theta \in (\pi /2,\pi ]$, then ${\sin }^{-1}(\sin 2\theta )=\pi -2\theta$, leading to $\theta =\pi -2\theta \Rightarrow 3\theta =\pi \Rightarrow \theta =\pi /3$.
7. If $2\theta \in [-\pi ,-\pi /2)$, then ${\sin }^{-1}(\sin 2\theta )=-\pi -2\theta$, giving $\theta =-\pi -2\theta \Rightarrow 3\theta =-\pi \Rightarrow \theta =-\pi /3$.
8. All three values $\theta =-\pi /3,0,\pi /3$ lie inside $(-\pi /2,\pi /2)$, so they are admissible.
9. Convert back to $x$: $x=\tan \theta \Rightarrow x=-\sqrt{3},0,\sqrt{3}$.

Concept tested: solving equations with inverse trigonometric functions using substitution

Published 6 September 2026 · How this solution was prepared and checked