The number of elements in the set

$$S=\left\{\theta \in[0,2 \pi]: 3 \cos ^{4} \theta-5 \cos ^{2} \theta-2 \sin ^{6} \theta+2=0\right\}$$ is :

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