Question

The number of elements in the set $S = \\{ \theta \in [-4\pi, 4\pi] : 3 \cos^2(2\theta) + 6 \cos^2(\theta) - 10\cos(2\theta) + 5 = 0 \\}$ is______.

Answer 1

The correct answer is: 32

3 \cos^2(2\theta) + 6 \cos^2(\theta) - 10\cos(2\theta) + 5 = 0$$-\frac{10(1+cos2θ)}{2}

$3\cos^2(2\theta) + \cos^2(\theta) = 0$

$\cos^2(\theta) = 0 \quad \text{or} \quad \cos^2(\theta) =$ $-\frac{1}{3}$

As $\theta \in [0, \pi], \quad \cos(2\theta) = -\frac{1}{3} ⇒ 2 times$

$⇒$ \theta \in [-4\pi, 4\pi], \quad \cos(2\theta) = -\frac{1}{3}$$-\frac{1}{3} $⇒$ 16 times

Similarly, $\cos(2\theta) = 0$ $⇒ 16$ times

∴ Total is 32 solutions.