Question

If $\frac{1}{6} \sin \theta, \cos \theta, \tan \theta$ are in G.P., then the general solution of $\theta$ is

• A. $2n\pi \pm \frac{\pi}{3}, \text{n} \in \mathbb{Z}$
• B. $n\pi + \frac{\pi}{3}, \text{n} \in \mathbb{Z}$
• C. $n\pi + \frac{\pi}{4}, \text{n} \in \mathbb{Z}$
• D. $2n\pi \pm \frac{\pi}{6}, \text{n} \in \mathbb{Z}$

Hint:

For $a, b, c$ in G.P., $b^2 = ac$.

Answer 1

The Correct Option is A

Step 1: G.P. Condition
$b^2 = ac \implies \cos^2 \theta = \frac{1}{6} \sin \theta \cdot \tan \theta$.
Step 2: Simplify
$\cos^2 \theta = \frac{\sin^2 \theta}{6 \cos \theta} \implies 6 \cos^3 \theta = 1 - \cos^2 \theta$.
Let $t = \cos \theta \implies 6t^3 + t^2 - 1 = 0$.
Step 3: Solve
By trial, $t = 1/2$ is a root: $6(1/8) + 1/4 - 1 = 3/4 + 1/4 - 1 = 0$.
$\cos \theta = 1/2 \implies \theta = 2n\pi \pm \frac{\pi}{3}$.
Final Answer: (A)