Answer
See proof below.
Work Step by Step
$r(\theta) $ is continuous and differentiable on $(0,\displaystyle \frac{\pi}{2})$.
$r(\displaystyle \frac{1}{2})\approx-1.986\lt 0, \qquad r(1)\approx 5.058\gt 0$
so the Intermediate Value Theorem guarantees that there is a zero between $\displaystyle \frac{1}{2}$ and $1$.
Check whether $r$ is increasing over $(0,\displaystyle \frac{\pi}{2})$.
(If it is, the mentioned zero would be the only zero.)
$r'(\displaystyle \theta)=\sec\theta\tan\theta-(-3\theta^{-4})=\sec\theta\tan\theta+\frac{3}{\theta^{4}}$
$\sec\theta\tan\theta $ is positive on$ (0,\displaystyle \frac{\pi}{2}), \quad$ as is $\displaystyle \frac{3}{\theta^{4}}$.
It follows that $r'$ is always positive on$ (0,\displaystyle \frac{\pi}{2}),$ which means that $r $ is an increasing function on$ (0,\displaystyle \frac{\pi}{2}),$
So, $r$ has only one zero in $(0,\displaystyle \frac{\pi}{2})$
