The general solution of 4 θ= 3 θ is θ=·s.

Correct Answer is : 2nπ7,n∈Z2 n π/7, n Z72nπ,n∈Z

Correct Answer is : 2nπ7,n∈Z2 n π/7, n Z72nπ,n∈Z

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MHTCET 11 Apr 2026 Shift 2 Paper

Question : 105 of 150

Marks: +1, -0

\cos 4 \theta=\cos 3 \theta

\theta=\cdots

Solution:

Concept:

Use the standard result: if

\cos A = \cos B

, then

A = 2n\pi \pm B

, where

n \in \mathbb{Z}

Explanation:

Given:

\cos 4\theta = \cos 3\theta

Here,

A = 4\theta

and

B = 3\theta

So,

4\theta = 2n\pi \pm 3\theta

Taking the positive sign:

4\theta = 2n\pi + 3\theta \Rightarrow \theta = 2n\pi

Taking the negative sign:

4\theta = 2n\pi - 3\theta \Rightarrow 7\theta = 2n\pi \Rightarrow \theta = \frac{2n\pi}{7}

The values

\theta = 2n\pi

are already included in

\theta = \frac{2n\pi}{7}

when

n

is a multiple of

7

Therefore, the complete general solution is

\theta = \frac{2n\pi}{7}

n \in \mathbb{Z}

Answer:

Option A:

\theta = \frac{2n\pi}{7}

n \in \mathbb{Z}

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