Answer
$ks\left(\frac{r_2^{4}-r_1^{4}\cos(\theta)}{r_1^{4}r_2^{4}\sin^{2}(\theta)}\right)=0$
Work Step by Step
From the previous part if follows:
$$\frac{dR}{d\theta}=ks\left(\frac{r_2^{4}-r_1^{4}\cos(\theta)}{r_1^{4}r_2^{4}\sin^{2}(\theta)}\right)$$
so:
$$ks\left(\frac{r_2^{4}-r_1^{4}\cos(\theta)}{r_1^{4}r_2^{4}\sin^{2}(\theta)}\right)=0$$
