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.

Given $f\left(x\right)= x^2-x\sin \left(x\right)-\cos \left(x\right)$. You are studying  $x^2-x\sin \left(x\right)-\cos \left(x\right)=0$ so when the function intersects x-axis. Study its monotony by studying its derivative $$f'\left(x\right)=2x-x\cos \left(x\right)$$ You will find that the derivative is positive for $x>0$ and $f\left(0\right)=-1$ so $f\left(x\right)$ will intersect x-axis only once in $\left(0,+∞\right)$.
