\(\sin x + \sin 5x = 2\sin 3x\cos 2x\). So the equation becomes \(2\sin 3x\cos 2x = \sin 3x\), or \(\sin 3x(2\cos 2x - 1) = 0\).

• \(\sin 3x = 0\): \(x = 0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi\). That's 4 solutions.
• \(\cos 2x = \frac{1}{2}\): \(2x = \frac{\pi}{3}\) or \(\frac{5\pi}{3}\), so \(x = \frac{\pi}{6}, \frac{5\pi}{6}\). That's 2 solutions.

Total = 6.

Answer: (B) 6