Answer
See proof below.
Work Step by Step
$r(\theta) $ is continuous and differentiable on $(-\infty,\infty)$.
$r(\displaystyle \frac{\pi}{2})=-2.48\lt 0, \qquad r(\frac{\pi}{2})\approx 3.81\gt 0$
so the Intermediate Value Theorem guarantees that there is a zero between $-\displaystyle \frac{\pi}{2}$ and $\displaystyle \frac{\pi}{2}$.
Check whether $r$ is increasing over $(-\infty,\infty)$.
(If it is, the mentioned zero would be the only zero.)
$ r'(\theta)=2-2\cos\theta(-\sin\theta)=2+\sin 2\theta$
Since $\sin(2\theta)$ can not be less than $-1$, it follows that $r'$ is always positive, which means that $r $ is an increasing function.
So, $r$ has only one zero in $(-\infty,\infty)$
