Question

The number of solutions of $\sin \, 3x = \cos \, 2x$, in the interval $\left( \frac{\pi}{2} , \pi \right)$ is :

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The Correct Option is A

To solve the equation \(\sin 3x = \cos 2x\) in the interval \(\left( \frac{\pi}{2}, \pi \right)\), we start by using trigonometric identities to simplify and solve the equation. 

First, recall the trigonometric identity:

\(\cos \theta = \sin \left( \frac{\pi}{2} - \theta \right)\)

Apply this identity to \(\cos 2x\):

\(\cos 2x = \sin \left( \frac{\pi}{2} - 2x \right)\)

Therefore, the equation becomes:

\(\sin 3x = \sin \left( \frac{\pi}{2} - 2x \right)\)

For the equation \(\sin A = \sin B\), the solution is given by:

\(A = n\pi + (-1)^n B\) for \(n \in \mathbb{Z}\)

Applying this to our equation:

1. \(3x = n\pi + (-1)^n \left( \frac{\pi}{2} - 2x \right)\)

Now, we need to solve for \(x\) and find solutions within the interval \(\left( \frac{\pi}{2}, \pi \right)\).

Let's consider both cases:

1. For \(n=0\):
   • \(3x = \frac{\pi}{2} - 2x\)
   • Solve for \(x\):
   • \(5x = \frac{\pi}{2}\)
   • \(x = \frac{\pi}{10}\) which is outside the interval \(\left( \frac{\pi}{2}, \pi \right)\)
2. For \(n=1\):
   • \(3x = \pi - \left( \frac{\pi}{2} - 2x \right)\)
   • Simplifying: \(3x = \pi - \frac{\pi}{2} + 2x\)
   • \(x = \frac{\pi}{2}\)
   • This value doesn't lie strictly inside the interval \(\left( \frac{\pi}{2}, \pi \right)\)

Upon checking, the valid solution which lies in the interval \(\left( \frac{\pi}{2}, \pi \right)\) occurs only for \(n=2\):

1. For \(n=2\): Solve \(3x = 2\pi - \left( \frac{\pi}{2} - 2x \right)\)
   • \(3x = 2\pi - \frac{\pi}{2} + 2x\)
   • Simplifying gives: \(x = \frac{3\pi}{5}\)
   • This solution lies within the interval \(\left( \frac{\pi}{2}, \pi \right)\)

Hence, the number of solutions in the interval \(\left( \frac{\pi}{2}, \pi \right)\) is \(1\).