The sum of all values of x in [0, 2pi],for wh | vedclass

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(D) Given equation: $(\sin x + \sin 4x) + (\sin 2x + \sin 3x) = 0$Using the sum-to-product formula $\sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{

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(D) Given equation: $(\sin x + \sin 4x) + (\sin 2x + \sin 3x) = 0$Using the sum-to-product formula $\sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$:$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 $1$: $\sin \frac{5x}{2} = 0 \Rightarrow \frac{5x}{2} = n\pi \Rightarrow x = \frac{2n\pi}{5}$. For $x \in [0, 2\pi]$,$x \in \{0, \frac{2\pi}{5}, \frac{4\pi}{5}, \frac{6\pi}{5}, \frac{8\pi}{5}, 2\pi\}$.Case $2$: $\cos x = 0 \Rightarrow x = \frac{\pi}{2}, \frac{3\pi}{2}$.Case $3$: $\cos \frac{x}{2} = 0 \Rightarrow \frac{x}{2} = \frac{\pi}{2} \Rightarrow x = \pi$.Sum of all values $= (0 + \frac{2\pi}{5} + \frac{4\pi}{5} + \frac{6\pi}{5} + \frac{8\pi}{5} + 2\pi) + (\frac{\pi}{2} + \frac{3\pi}{2}) + \pi = 6\pi + 2\pi + \pi = 9\pi$.

The sum of all values of $x$ in $[0, 2\pi]$,for which $\sin x + \sin 2x + \sin 3x + \sin 4x = 0$,is equal to: (in $\pi$)

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