Answer
See proof below.
Work Step by Step
$r(\theta) $ is continuous and differentiable on $(-\infty,\infty)$.
$r(0)=-8\lt 0, \qquad r(3\pi)\approx 1.423\gt 0$
so the Intermediate Value Theorem guarantees that there is a zero between $0$ and $ 3\pi$.
Check whether $r$ is increasing over $(-\infty,\infty)$.
(If it is, the mentioned zero would be the only zero.)
$r'(\displaystyle \theta)=1+2\sin(\frac{\theta}{3})\cdot\cos(\frac{\theta}{3})\cdot\frac{1}{3}=1+\frac{1}{3}\sin(\frac{2\theta}{3})$
Since $\displaystyle \sin(\frac{2\theta}{3})$ 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 $(-\infty,\infty)$.
