Question

The number of solutions of the equation $4 \cos 2\theta \cos 3\theta = \sec \theta$ in the interval $[0, 2\pi]$ is

• A. 12
• B. 8
• C. 16
• D. 4

Hint:

Convert complex trigonometric equations to sums of cosines and solve within given interval considering domain restrictions.

Answer 1

The Correct Option is A

Rewrite left side using product-to-sum: \[ 4 \cos 2\theta \cos 3\theta = 2 \left[ \cos(5\theta) + \cos(\theta) \right]. \] Given equation becomes \[ 2 [ \cos 5\theta + \cos \theta ] = \sec \theta = \frac{1}{\cos \theta}. \] Multiply both sides by $\cos \theta$ (exclude values where $\cos \theta=0$): \[ 2 \cos \theta \cos 5\theta + 2 \cos^2 \theta = 1. \] Use product-to-sum again for $\cos \theta \cos 5\theta$: \[ \cos \theta \cos 5\theta = \frac{1}{2} [\cos 6\theta + \cos 4\theta]. \] So, \[ 2 \times \frac{1}{2} [\cos 6\theta + \cos 4\theta] + 2 \cos^2 \theta = 1, \] which simplifies to \[ \cos 6\theta + \cos 4\theta + 2 \cos^2 \theta = 1. \] Use $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$, \[ \cos 6\theta + \cos 4\theta + 1 + \cos 2\theta = 1, \] so \[ \cos 6\theta + \cos 4\theta + \cos 2\theta = 0. \] Using sum-to-product formulas and considering the domain $[0, 2\pi]$, find the number of solutions to this trigonometric equation. The count is 12 solutions.