Show that $x^2 = \sin (x)$ has exactly one positive solution I want to show that $x^2 = \sin(x)$ has exactly one positive solution.
We know that $x^2 - \sin(x)$ only has roots in the segment $[-1,1]$ and we also know that $x=0$ is a root.
How do I show that there exists exactly one root on $(0,\infty)$?

Hint As you point out, outside the  $[-1,1]$ interval there cannot be solutions, as for $x>1$, $x^2 > \max \sin(x)$.
The existence of a positive solution can be proved by noting that for sufficienty small $x$, $x^2 < \sin x$, while on the other hand $ \sin (1) < 1^2 =1 $: they must then "cross" somewhere in between, very loosely speaking.
Now, $x^2$ is convex, while $\sin$ is concave over the same $(0,1)$ interval, and that should help proving there is only one positive root.
