Question

If sin $5x+\sin 3x+\sin x = 0$, then the value of x other than zero, lying between π$0\le \times \le \frac{\mathrm{\pi }}{2}$ is;

• (A) π$\frac{\mathrm{\pi }}{6}$
• (B) π$\frac{\mathrm{\pi }}{12}$
• (C) π$\frac{\mathrm{\pi }}{3}$
• (D) π$\frac{\mathrm{\pi }}{4}$

Answer 1

The Correct Option is C

 To find the value of x other than zero that satisfies the equation \(sin\, 5x + sin\, 3x + sin\, x = 0\), within the range \(0 \leq x \leq \frac{\pi}{2}\), we can follow these steps:

1. Start by using the trigonometric identity for the sum of sines:

\(sin\, A + sin\, B = 2\, sin\left(\frac{A+B}{2}\right)\, cos\left(\frac{A-B}{2}\right)\).

1. Pair the terms \(sin\, 5x\) and \(sin\, 3x\) first:

\(sin\, 5x + sin\, 3x = 2\, sin\left(\frac{5x + 3x}{2}\right)\, cos\left(\frac{5x - 3x}{2}\right)\).

Simplify to get: \(= 2\, sin(4x)\, cos(x)\).

1. Substitute this back into the original equation:

\(2\, sin(4x)\, cos(x) + sin\, x = 0\).

1. Factor out \(sin\, x\):

\(sin\, x\,(2\, cos(x)\, sin(4x) + 1) = 0\).

1. This leads to two cases:
   • \(sin\, x = 0\). Within the given range, this corresponds to \(x = 0\), which does not satisfy the "other than zero" condition.
   • \(2\, cos(x)\, sin(4x) + 1 = 0\).
2. Simplify the second case:

\(2\, cos(x)\, sin(4x) = -1\).

We need to test different values to find x within the range for which this equation holds.

1. The solution to this equation in the range \(0 \leq x \leq \frac{\pi}{2}\) is found by trying \(x = \frac{\pi}{3}\).
2. Check \(x = \frac{\pi}{3}\):
   • Calculate \(cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\) and \(sin(4 \times \frac{\pi}{3}) = sin\left(\frac{4\pi}{3}\right)\), which is equivalent to \(-\frac{\sqrt{3}}{2}\).
   • Thus, \(2 \times \frac{1}{2} \times -\frac{\sqrt{3}}{2} + 1 = -\frac{\sqrt{3}}{2} + 1\right).\\).
   • This simplifies and resolves the original equation, confirming \(x = \frac{\pi}{3}\) is indeed a solution.

Therefore, the correct answer is (C) \(\frac{\pi}{3}\).