The original equation is: $3 \cos ^4 \theta-5 \cos ^2 \theta-2 \sin ^6 \theta+2=0$ We can re-arrange terms and use trigonometric identity ($\cos^2 \theta + \sin^2 \theta = 1$) to rewrite this as : $3 \cos ^4 \theta - 3 \cos ^2 \theta + 2 \sin ^2 \theta - 2 \sin ^6 \theta = 0$ Factoring out $\cos^2 \theta$ and $\sin^2 \theta$ we get : $3 \cos ^2 \theta (\cos ^2 \theta - 1) + 2 \sin ^2 \theta (\sin ^4 \theta - 1) = 0$ Then we substitute $1 - \cos^2 \theta$ for $\sin^2 \theta$ (again using the same trigonometric identity), which simplifies to : $-3 \cos ^2 \theta \sin ^2 \theta + 2 \sin ^2 \theta(1 + \sin ^2 \theta)\cos ^2 \theta = 0$ This can be re-written as : $\sin ^2 \theta \cos ^2 \theta(2 + 2 \sin ^2 \theta - 3) = 0$ Simplifying it to : $\sin ^2 \theta \cos ^2 \theta(2 \sin ^2 \theta - 1) = 0$ We now have 3 separate conditions to solve for : - $\sin ^2 \theta = 0$, - $\cos ^2 \theta = 0$, and - $2 \sin ^2 \theta - 1 = 0$. For $\sin ^2 \theta = 0$, the solutions are $\theta=\{0, \pi, 2 \pi\}$ (3 solutions). For $\cos ^2 \theta = 0$, the solutions are $\theta=\left\{\frac{\pi}{2}, \frac{3 \pi}{2}\right\}$ (2 solutions). For $2 \sin ^2 \theta - 1 = 0$ (which simplifies to $\sin ^2 \theta = 1/2$), the solutions are $\theta=\left\{\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}\right\}$ (4 solutions). In total, this gives 3+2+4 = 9 solutions.