If 0 leq x leq 2 pi,then the number of real v | vedclass

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(A) Given equation: $\sin x + \sin 2x + \sin 3x + \sin 4x = 0$. Grouping terms: $(\sin 4x + \sin x) + (\sin 3x + \sin 2x) = 0$. Using the formula $\si

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$9$

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(A) Given equation: $\sin x + \sin 2x + \sin 3x + \sin 4x = 0$. Grouping terms: $(\sin 4x + \sin x) + (\sin 3x + \sin 2x) = 0$. Using the formula $\sin C + \sin D = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})$: $2 \sin(\frac{5x}{2}) \cos(\frac{3x}{2}) + 2 \sin(\frac{5x}{2}) \cos(\frac{x}{2}) = 0$. $2 \sin(\frac{5x}{2}) [\cos(\frac{3x}{2}) + \cos(\frac{x}{2})] = 0$. Using $\cos C + \cos D = 2 \cos(\frac{C+D}{2}) \cos(\frac{C-D}{2})$: $2 \sin(\frac{5x}{2}) [2 \cos x \cos(\frac{x}{2})] = 0$. $4 \sin(\frac{5x}{2}) \cos x \cos(\frac{x}{2}) = 0$. Case $1$: $\sin(\frac{5x}{2}) = 0 \implies \frac{5x}{2} = n \pi \implies x = \frac{2n \pi}{5}$. For $0 \leq x \leq 2 \pi$,$x \in \{0, \frac{2 \pi}{5}, \frac{4 \pi}{5}, \frac{6 \pi}{5}, \frac{8 \pi}{5}, 2 \pi\}$. Case $2$: $\cos x = 0 \implies x = \frac{\pi}{2}, \frac{3 \pi}{2}$. Case $3$: $\cos(\frac{x}{2}) = 0 \implies \frac{x}{2} = \frac{\pi}{2} \implies x = \pi$. Combining all unique values: $\{0, \frac{2 \pi}{5}, \frac{\pi}{2}, \frac{4 \pi}{5}, \pi, \frac{6 \pi}{5}, \frac{3 \pi}{2}, \frac{8 \pi}{5}, 2 \pi\}$. Total number of values is $9$.

If $0 \leq x \leq 2 \pi$,then the number of real values of $x$ which satisfy the equation $\sin x + \sin 2x + \sin 3x + \sin 4x = 0$ is

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