Test Index

MHTCET 16 Apr 2026 Shift 1 Paper

Question : 105 of 150

Marks: +1, -0

0 \leq x \leq \frac{\pi}{2}

x

\sin x-\sin 2 x+\sin 3 x=0

Solution:

Concept:

Use trigonometric identities to simplify the equation, then find the number of distinct roots in the given interval.

Explanation:

Given:

\sin x - \sin 2x + \sin 3x = 0

Rewrite as

\sin x + \sin 3x - \sin 2x = 0

Using

\sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}

, we get:

\sin x + \sin 3x = 2 \sin 2x \cos x

So the equation becomes

2 \sin 2x \cos x - \sin 2x = 0

Factorize:

\sin 2x (2 \cos x - 1) = 0

Hence,

\sin 2x = 0

2 \cos x - 1 = 0

For

0 \leq x \leq \frac{\pi}{2}

\sin 2x = 0

gives

x = 0, \frac{\pi}{2}

2 \cos x - 1 = 0

gives

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

, so

x = \frac{\pi}{3}

The distinct values of

x

are

0, \frac{\pi}{3}, \frac{\pi}{2}

Thus, the number of values is

3

Answer:

Option C:

3

Go to Question:

Prev Question

Next Question