Question:medium

If \(\sin\theta\neq 0\) and \[ \frac{1}{2}\sin\theta,\quad \cos\theta,\quad \cot\theta \] are in geometric progression, then the number of values of \(\theta\) lying in the interval \((-2\pi,2\pi)\) is

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Whenever three terms are given in G.P., immediately use \[ b^2=ac. \] After obtaining the trigonometric equation, solve each case separately and carefully count all solutions in the specified interval.
Updated On: Jul 9, 2026
  • \(5\)
  • \(7\)
  • \(10\)
  • \(8\) \bigskip
Show Solution

The Correct Option is D

Solution and Explanation

Concept: Use the geometric progression condition \(b^2 = ac\) to form an equation, solve for \(\cos\theta\) using algebraic manipulation, and then find all solutions in the given interval.

Step 1:
Apply the G.P. condition. Given \(\frac{1}{2}\sin\theta, \cos\theta, \cot\theta\) are in G.P.: \[ \cos^2\theta = \left(\frac{1}{2}\sin\theta\right)(\cot\theta) = \frac{1}{2}\sin\theta \cdot \frac{\cos\theta}{\sin\theta} = \frac{1}{2}\cos\theta. \] Rearrange: \[ \cos^2\theta - \frac{1}{2}\cos\theta = 0 \implies \cos\theta\left(\cos\theta - \frac{1}{2}\right) = 0. \]

Step 2:
Solve for \(\theta\) in \((-2\pi, 2\pi)\). Case 1: \(\cos\theta = 0 \implies \theta = \frac{\pi}{2} + n\pi\). In \((-2\pi, 2\pi)\): \(\theta = -\frac{3\pi}{2}, -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}\) (4 solutions). Case 2: \(\cos\theta = \frac{1}{2} \implies \theta = 2n\pi \pm \frac{\pi}{3}\). In \((-2\pi, 2\pi)\): \(\theta = -\frac{5\pi}{3}, -\frac{\pi}{3}, \frac{\pi}{3}, \frac{5\pi}{3}\) (4 solutions).

Step 3:
Count total solutions and write the final answer. Total solutions = 4 + 4 = 8. \[ \boxed{8} \]
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