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(A) Given equation: $\sin x - \sin 2x + \sin 3x = 2 \cos^2 x - 2 \cos x$Rearranging terms: $(\sin x + \sin 3x) - \sin 2x = 2 \cos x(\cos x - 1)$Using

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(A) Given equation: $\sin x - \sin 2x + \sin 3x = 2 \cos^2 x - 2 \cos x$Rearranging terms: $(\sin x + \sin 3x) - \sin 2x = 2 \cos x(\cos x - 1)$Using the sum-to-product formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:$2 \sin 2x \cos x - \sin 2x = 2 \cos x(\cos x - 1)$$\sin 2x(2 \cos x - 1) = 2 \cos x(\cos x - 1)$$2 \sin x \cos x(2 \cos x - 1) = 2 \cos x(\cos x - 1)$$2 \cos x [\sin x(2 \cos x - 1) - (\cos x - 1)] = 0$$2 \cos x [2 \sin x \cos x - \sin x - \cos x + 1] = 0$$2 \cos x [\sin x(2 \cos x - 1) - 1(\cos x - 1)] = 0$Actually,simplifying the original equation:$\sin x + \sin 3x = 2 \sin 2x \cos x$So,$2 \sin 2x \cos x - \sin 2x = 2 \cos^2 x - 2 \cos x$$\sin 2x(2 \cos x - 1) = 2 \cos x(\cos x - 1)$$2 \sin x \cos x(2 \cos x - 1) = 2 \cos x(\cos x - 1)$Case $1$: $\cos x = 0 \Rightarrow x = \frac{\pi}{2}$ (Valid in $(0, \pi)$)Case $2$: $\sin x(2 \cos x - 1) = \cos x - 1$$2 \sin x \cos x - \sin x = \cos x - 1$$\sin 2x - \sin x - \cos x + 1 = 0$For $x \in (0, \pi)$,testing values:If $x = \frac{\pi}{6}$,$\sin \frac{\pi}{3} - \sin \frac{\pi}{6} - \cos \frac{\pi}{6} + 1 = \frac{\sqrt{3}}{2} - \frac{1}{2} - \frac{\sqrt{3}}{2} + 1 = \frac{1}{2} 
eq 0$If $x = \frac{5\pi}{6}$,$\sin \frac{5\pi}{3} - \sin \frac{5\pi}{6} - \cos \frac{5\pi}{6} + 1 = -\frac{\sqrt{3}}{2} - \frac{1}{2} + \frac{\sqrt{3}}{2} + 1 = \frac{1}{2} 
eq 0$Re-evaluating: $\sin x - \sin 2x + \sin 3x = 2 \cos^2 x - 2 \cos x$$2 \sin 2x \cos x - \sin 2x = 2 \cos x(\cos x - 1)$$\sin 2x(2 \cos x - 1) = 2 \cos x(\cos x - 1)$$2 \sin x \cos x(2 \cos x - 1) = 2 \cos x(\cos x - 1)$$2 \cos x [\sin x(2 \cos x - 1) - (\cos x - 1)] = 0$$2 \cos x [2 \sin x \cos x - \sin x - \cos x + 1] = 0$$2 \cos x [\sin x(2 \cos x - 1) - 1(\cos x - 1)] = 0$This leads to $\cos x = 0$ or $2 \sin x \cos x - \sin x - \cos x + 1 = 0$$2 \sin x \cos x - \sin x - \cos x + 1 = (2 \cos x - 1)(\sin x - 0.5) + 0.5 = 0$Actually,the solutions are $x = \frac{\pi}{2}$ and $x = \frac{\pi}{6}, \frac{5\pi}{6}$ is incorrect. The only solution is $x = \frac{\pi}{2}$.

Number of solutions of the equation $\sin x - \sin 2x + \sin 3x = 2 \cos^2 x - 2 \cos x$ in the interval $(0, \pi)$ is

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