Question

If $0 \le x < \frac{\pi}{2}$ , then the number of values of $ x$ for which $sin \,x-sin\,2x+sin\,3x = 0$, is

• A. 2
• B. 1
• C. 3
• D. 4

Answer 1

The Correct Option is A

To solve the equation \(\sin x - \sin 2x + \sin 3x = 0\) for \(0 \le x < \frac{\pi}{2}\), we start by using trigonometric identities.

1. Let's use the sine addition formulas:
   • \(\sin 2x = 2 \sin x \cos x\)
   • \(\sin 3x = 3 \sin x - 4 \sin^3 x\)
2. Substitute these identities into the given equation: 

\[\sin x - (2 \sin x \cos x) + (3 \sin x - 4 \sin^3 x) = 0\]

1. Simplify the equation:
   • Combine terms: \(\sin x + 3 \sin x - 2 \sin x \cos x - 4 \sin^3 x = 0\)
   • Factor out \(\sin x\): \(\sin x (4 - 2 \cos x - 4 \sin^2 x) = 0\)
2. This gives two cases:
   • Case 1: \(\sin x = 0\)
     • In the interval \(0 \le x < \frac{\pi}{2}\), the only solution is \(x = 0\).
   • Case 2: \(4 - 2 \cos x - 4 \sin^2 x = 0\)
     • Use the identity \(\sin^2 x = 1 - \cos^2 x\):
     • Substitute: \(4 - 2 \cos x - 4(1 - \cos^2 x) = 0\)
     • Simplify: \(4 - 2 \cos x - 4 + 4 \cos^2 x = 0\)
     • Write as a quadratic in \(\cos x\): \(4 \cos^2 x - 2 \cos x = 0\)
     • Factor: \(2 \cos x (2 \cos x - 1) = 0\)
     • This gives \(\cos x = 0\) or \(\cos x = \frac{1}{2}\).
     • In the interval \(0 \le x < \frac{\pi}{2}\), \(x = \frac{\pi}{3}\) and \(x = \frac{\pi}{2}\) solutions are not valid as \(x = \frac{\pi}{2}\) is outside the open interval on this end.

Thus, the two valid solutions for \(x\) in the range \(0 \le x < \frac{\pi}{2}\) are \(x = 0\) and \(x = \frac{\pi}{3}\).

Therefore, the number of values of \(x\) satisfying the equation is 2.