Using sum-to-product: \((\sin x + \sin 3x) - \sin 2x = 0\), which gives \(2\sin 2x \cos x - \sin 2x = 0\), or \(\sin 2x(2\cos x - 1) = 0\). Either \(\sin 2x = 0\) or \(\cos x = \frac{1}{2}\). For \(\sin 2x = 0\): \(2x = 0, \pi, 2\pi, \ldots\), giving \(x = 0\) (boundary, included) or \(x = \frac{\pi}{2}\) (boundary, excluded). So one solution \(x=0\). For \(\cos x = \frac{1}{2}\): \(x = \frac{\pi}{3}\) in the given range. Total solutions = 2.