For n in mathbb{Z},the general solution of th | vedclass

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(B) Given equation: $\sin x - \sqrt{3} \cos x + 4 \sin 2x - 4 \sqrt{3} \cos 2x + \sin 3x - \sqrt{3} \cos 3x = 0$. Divide by $2$ to use the identity $\

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$\frac{n \pi}{2} + \frac{\pi}{6}$

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(B) Given equation: $\sin x - \sqrt{3} \cos x + 4 \sin 2x - 4 \sqrt{3} \cos 2x + \sin 3x - \sqrt{3} \cos 3x = 0$. Divide by $2$ to use the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$: $2[\frac{1}{2} \sin x - \frac{\sqrt{3}}{2} \cos x] + 8[\frac{1}{2} \sin 2x - \frac{\sqrt{3}}{2} \cos 2x] + 2[\frac{1}{2} \sin 3x - \frac{\sqrt{3}}{2} \cos 3x] = 0$. This simplifies to: $2 \sin(x - \frac{\pi}{3}) + 8 \sin(2x - \frac{\pi}{3}) + 2 \sin(3x - \frac{\pi}{3}) = 0$. Divide by $2$: $\sin(x - \frac{\pi}{3}) + \sin(3x - \frac{\pi}{3}) + 4 \sin(2x - \frac{\pi}{3}) = 0$. Using $\sin C + \sin D = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})$: $2 \sin(2x - \frac{\pi}{3}) \cos(x) + 4 \sin(2x - \frac{\pi}{3}) = 0$. $2 \sin(2x - \frac{\pi}{3}) [\cos x + 2] = 0$. Since $\cos x + 2 
eq 0$ for any real $x$,we must have $\sin(2x - \frac{\pi}{3}) = 0$. Thus,$2x - \frac{\pi}{3} = n \pi$,which gives $2x = n \pi + \frac{\pi}{3}$,or $x = \frac{n \pi}{2} + \frac{\pi}{6}$.

For $n \in \mathbb{Z}$,the general solution of the trigonometric equation $\sin x - \sqrt{3} \cos x + 4 \sin 2x - 4 \sqrt{3} \cos 2x + \sin 3x - \sqrt{3} \cos 3x = 0$ is

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