If $\theta \in [-2\pi, 2\pi]$, then the number of solutions of $2\sqrt{2}\cos^2\theta + (2-\sqrt{6})\cos\theta – \sqrt{3} = 0$ is equal to:
- (1) 12
- (2) 6
- (3) 8
- (4) 10
Solution:
Factorize the given quadratic equation in terms of $\cos\theta$:
$$2\sqrt{2}\cos^2\theta + 2\cos\theta – \sqrt{6}\cos\theta – \sqrt{3} = 0$$
$$\therefore 2\cos\theta(\sqrt{2}\cos\theta + 1) – \sqrt{3}(\sqrt{2}\cos\theta + 1) = 0$$
$$\therefore (2\cos\theta – \sqrt{3})(\sqrt{2}\cos\theta + 1) = 0$$
$$\therefore \cos\theta = \frac{\sqrt{3}}{2} \quad \text{or} \quad \cos\theta = -\frac{1}{\sqrt{2}}$$
We need to count solutions in the interval $\theta \in [-2\pi, 2\pi]$.
Case 1: $\cos\theta = \frac{\sqrt{3}}{2}$
In the interval $[0, 2\pi]$, solutions are $\frac{\pi}{6}, \frac{11\pi}{6}$ (2 solutions).
Since the range is $[-2\pi, 2\pi]$ (which covers two full periods), there are $2 \times 2 = 4$ solutions.
Case 2: $\cos\theta = -\frac{1}{\sqrt{2}}$
In the interval $[0, 2\pi]$, solutions are $\frac{3\pi}{4}, \frac{5\pi}{4}$ (2 solutions).
Similarly, in $[-2\pi, 2\pi]$, there are $2 \times 2 = 4$ solutions.
Total number of solutions = $4 + 4 = 8$.
Ans. (3)