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MHTCET 20 Apr 2026 Shift 1 Paper

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Question : 106 of 150

Marks: +1, -0

The general solution of

\cos x+\sin x=\cos 2 x+\sin 2 x

x=n p \pi

x=\frac{n q \pi}{3}+\frac{\pi}{6}

n \in Z

p: q=

1 : 1
1 : 2
2 : 3
2 : 1

Solution:

Concept:

Use the identity

\sin \theta + \cos \theta = \sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right)

and the general solution of

\sin A = \sin B

.

Explanation:

Rewrite both sides using the identity.

\cos x + \sin x = \sqrt{2}\sin\left(x + \frac{\pi}{4}\right)

\cos 2x + \sin 2x = \sqrt{2}\sin\left(2x + \frac{\pi}{4}\right)

Thus,

\sin\left(x + \frac{\pi}{4}\right) = \sin\left(2x + \frac{\pi}{4}\right)

.

For

\sin A = \sin B

, either

A = B + 2n\pi

or

A = \pi - B + 2n\pi

.

First case:

x + \frac{\pi}{4} = 2x + \frac{\pi}{4} + 2n\pi

This gives

x = -2n\pi

, which is same as

x = 2n\pi

.

Hence,

p = 2

.

Second case:

x + \frac{\pi}{4} = \pi - \left(2x + \frac{\pi}{4}\right) + 2n\pi

Simplifying gives

3x = \frac{\pi}{2} + 2n\pi

.

So,

x = \frac{\pi}{6} + \frac{2n\pi}{3}

.

Comparing with

x = \frac{nq\pi}{3} + \frac{\pi}{6}

, we get

q = 2

.

Therefore,

p:q = 2:2 = 1:1

.

Answer:

1:1

, i.e. Option A.

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