The number of elements in the set S = [0,2] : 34 - 52 - 26 + 2 = 0 is:

Question

The number of elements in the set $S = \{	heta\in[0,2\pi] : 3\cos4	heta - 5\cos2	heta - 2\sin6	heta + 2 = 0\}$ is:

Accepted Answer

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.

Solutions · 1