QJEEM23Maef6a: JEE Main 2023 Mathematics - Class 11 JEEM23Maef6a Solution: The number of elements in the set is :

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.