If 1/6 θ, θ, θ are in G.P., then the general solution of θ is

Correct Answer is : 2nπ±π3,n∈Z2 {n} π ± π/3, {n} {Z}2nπ±3π,n∈Z

Correct Answer is : 2nπ±π3,n∈Z2 {n} π ± π/3, {n} {Z}2nπ±3π,n∈Z

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MHTCET 22 Apr 2025 Shift 2 Paper

Question : 106 of 150

Marks: +1, -0

\frac{1}{6} \sin \theta, \cos \theta, \tan \theta

\theta

Solution:

Concept:

For three terms in G.P., the square of the middle term equals the product of the other two terms.

Explanation:

Given that

\frac{1}{6}\sin\theta

\cos\theta

\tan\theta

are in G.P.

Using the G.P. condition, we get:

\cos^2\theta = \frac{1}{6}\sin\theta \cdot \tan\theta

Since

\tan\theta = \frac{\sin\theta}{\cos\theta}

, we have:

\cos^2\theta = \frac{\sin^2\theta}{6\cos\theta}

\Rightarrow 6\cos^3\theta = \sin^2\theta

Using

\sin^2\theta = 1 - \cos^2\theta

6\cos^3\theta = 1 - \cos^2\theta

\Rightarrow 6\cos^3\theta + \cos^2\theta - 1 = 0

Factorizing:

\left(\cos\theta - \frac{1}{2}\right)\left(6\cos^2\theta + 4\cos\theta + 2\right) = 0

The quadratic factor has no real roots, so:

\cos\theta = \frac{1}{2}

Hence, the general solution is:

\theta = 2n\pi \pm \frac{\pi}{3}, \quad n \in \mathbb{Z}

Answer:

2n\pi \pm \frac{\pi}{3}, \ n \in \mathbb{Z}

Therefore, the correct option is A.

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