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(A) Given equation: $\sin x - \sin 2x + \sin 3x = 0$Using the sum-to-product formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:$(\

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$2$

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(A) Given equation: $\sin x - \sin 2x + \sin 3x = 0$Using the sum-to-product formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:$(\sin 3x + \sin x) - \sin 2x = 0$$2 \sin 2x \cos x - \sin 2x = 0$$\sin 2x (2 \cos x - 1) = 0$This implies $\sin 2x = 0$ or $\cos x = \frac{1}{2}$.For $\sin 2x = 0$,$2x = n\pi$,so $x = \frac{n\pi}{2}$. Given $0 \le x For $\cos x = \frac{1}{2}$,$x = \frac{\pi}{3}$ (since $0 \le x The values of $x$ are $0$ and $\frac{\pi}{3}$.Thus,the number of values is $2$.

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

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