Prove that a transcendental function has exactly one root on a given interval

Horizontal and vertical asymptotes of polar curve $r = \theta/(\pi - \theta) \, , \, \in[0,\pi]$

I need to show that the function

$$x^{\prime}(\theta) = \frac{\cos(\theta)\pi-\theta\sin(\theta)\pi+\theta^2\sin(\theta)}{(\pi - \theta)^2}$$

has exactly one solution on the interval $\theta \in [0,\pi)$. Is this possible to show rigorously ?

My attempt below (not with much rigour, I am ashamed to admit):

We know that the function has at least one root since $x'(0)>0$ and $x'(\pi/2)<0$. Now my idea was to find the maximum of the function, and from there prove that the function was strictly increasing before this point, and strictly decreasing after this point. Alas, I can not prove that the function only has one extremum on the interval, and I am having difficulties showing that it is strictly increasing and decreasing.

(Although finding a (the?) maximum is easy, by using Newton/Raphson approximation with the initial condition $x_0=\pi/3$.)

Is this a correct way of taking a stab at the problem, or is there any other smarter and more rigorous way of dealing with this problem?

-
First off, you can ignore the denominator; since it is always nonzero in the interval you're considering it cannot possibly change where the roots are. Now, plotting the numerator shows that it is not strictly decreasing, but it might be possible to show, for example, that it is strictly decreasing on $[0,\pi/2]$ and always negative on $[\pi/2, \pi)$. – Henning Makholm Jan 22 '12 at 18:38

You can rewrite the equation as $$\theta-\frac{\theta^2}{\pi}=\cot \theta\ .$$ On the interval $(\dfrac\pi2,\pi)$, $\cot\theta<0$ while $\theta-\dfrac{\theta^2}{\pi}>0$; we conclude that the only solutions are in the interval $(0,\dfrac\pi2)$ (since neither $0$ nor $\dfrac\pi2$ are solutions.) Finally, $\theta-\dfrac{\theta^2}{\pi}$ is increasing on $(0,\dfrac\pi2)$ while $\cot\theta$ is decreasing, proving that their graphs cross in at most one point.