If 4 θ- 2 θ=2, then θ=

Correct Answer is : nπ+π2,nπ5+π10,n∈Zn π+π/2, n π/5+π/10, n Znπ+2π,5nπ+10π,n∈Z

Correct Answer is : nπ+π2,nπ5+π10,n∈Zn π+π/2, n π/5+π/10, n Znπ+2π,5nπ+10π,n∈Z

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MHTCET 15 Apr 2026 Shift 2 Paper

Question : 107 of 150

Marks: +1, -0

\sec 4 \theta-\sec 2 \theta=2

\theta=

Solution:

Concept:

Convert the given secant equation into a polynomial in

\cos x

using identities, then solve for

\theta

Explanation:

Let

x = 2\theta

Then

\sec 4\theta - \sec 2\theta = 2

becomes

\sec 2x - \sec x = 2

Using

\sec A = \frac{1}{\cos A}

\frac{1}{\cos 2x} - \frac{1}{\cos x} = 2

\frac{\cos x - \cos 2x}{\cos x \cos 2x} = 2

\cos x - \cos 2x = 2\cos x \cos 2x

Put

c = \cos x

and use

\cos 2x = 2c^2 - 1

c - (2c^2 - 1) = 2c(2c^2 - 1)

4c^3 + 2c^2 - 3c - 1 = 0

Factorising:

(c+1)(4c^2 - 2c - 1) = 0

c = -1

c = \frac{1 \pm \sqrt{5}}{4}

Now,

\frac{1+\sqrt{5}}{4} = \cos \frac{\pi}{5}

and

\frac{1-\sqrt{5}}{4} = \cos \frac{3\pi}{5}

Hence all solutions for

x

combine to:

x = \frac{(2n+1)\pi}{5}, \quad n \in Z

Since

x = 2\theta

2\theta = \frac{(2n+1)\pi}{5}

\theta = \frac{(2n+1)\pi}{10}

This can be written as:

\theta = \frac{n\pi}{5} + \frac{\pi}{10}, \quad n \in Z

This form also includes values such as

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

for suitable

n

So the correct option is C.

Answer:

\theta = \frac{n\pi}{5} + \frac{\pi}{10}, \quad n \in Z

Option C.

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