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
$f $ is continuous and differentiable at $(0,+\infty). $
Assume that there exist $x_1>x_2>0$ such that
$$f (x_1)=f (x_2)=0=f (0) $$
then by Double Rolle Theorem, 
$$\exists c_1, c_2 \;\;:\;\; f'(c_1)=f'(c_2)=0$$
and
$$\exists c\in (c_1,c_2) \;\;:\;\; f''(c)=0$$
but
$$f''(x)=2+\sin(x)>0$$
