Prove that the equation $x^{2}-x\sin(x)-\cos(x)=0$ has only one root in the closed interval $(0,\infty)$. Here's the graph
(http://www.wolframalpha.com/input/?i=%28x%5E2%29-xsenx-cosx%3D0).
The part I'm having trouble with is proving that the root is unique. 
I can use the intermediate value theorem to find the interval where the root is, but from that I'm lost. 
I know you can look at the first or second derivative, but I don't know how to use that when the sign of sin and cos varies.
Thanks.

$f(x)=x^2-x \sin x -\cos(x)$ and $f'(x)=2x-x \cos x=x(2-\cos x)$. Clearly $2-\cos x>0$ and for $x>0$ we have that $f'(x)> 0$ $x>0$, therefore $f(x)$ is strctly increasing. now since $f(0)<0$ and $f(\infty)>0$, $f(x)$ just has one root on $x>0$.
This is a graphs of $f(x)$

and here is $f'(x)$

