If 0 leq x leq frac{pi}{2}, then the number o | vedclass

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(C) Given equation: $\sin x - \sin 2x + \sin 3x = 0$Rearrange the terms: $(\sin 3x + \sin x) - \sin 2x = 0$Using the sum-to-product formula $\sin A +

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$3$

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(C) Given equation: $\sin x - \sin 2x + \sin 3x = 0$Rearrange the terms: $(\sin 3x + \sin x) - \sin 2x = 0$Using the sum-to-product formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:$2 \sin(\frac{3x+x}{2}) \cos(\frac{3x-x}{2}) - \sin 2x = 0$$2 \sin 2x \cos x - \sin 2x = 0$Factor out $\sin 2x$: $\sin 2x (2 \cos x - 1) = 0$This implies $\sin 2x = 0$ or $2 \cos x - 1 = 0$.Case $1$: $\sin 2x = 0$. Since $0 \leq x \leq \frac{\pi}{2}$, then $0 \leq 2x \leq \pi$. Thus, $2x = 0$ or $2x = \pi$, which gives $x = 0$ or $x = \frac{\pi}{2}$.Case $2$: $2 \cos x - 1 = 0 \implies \cos x = \frac{1}{2}$. Since $0 \leq x \leq \frac{\pi}{2}$, $x = \frac{\pi}{3}$.The values of $x$ are $0, \frac{\pi}{3}, \frac{\pi}{2}$.Total number of values is $3$.

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

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