The equation sin 3theta = 4 sin theta sin 2th | vedclass

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(D) The given equation is $\sin 3	heta = 4 \sin 	heta \sin 2	heta \sin 4	heta$.Using $\sin 3	heta = 3 \sin 	heta - 4 \sin^3 	heta$,we get $3 \s

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$8$ real solutions

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(D) The given equation is $\sin 3	heta = 4 \sin 	heta \sin 2	heta \sin 4	heta$.Using $\sin 3	heta = 3 \sin 	heta - 4 \sin^3 	heta$,we get $3 \sin 	heta - 4 \sin^3 	heta = 4 \sin 	heta \sin 2	heta \sin 4	heta$.Case $1$: $\sin 	heta = 0$. In $0 \le 	heta \le \pi$,the solutions are $	heta = 0, \pi$ ($2$ solutions).Case $2$: $3 - 4 \sin^2 	heta = 4 \sin 2	heta \sin 4	heta$.Using $4 \sin^2 	heta = 2(1 - \cos 2	heta)$,we have $3 - 2(1 - \cos 2	heta) = 2(2 \sin 2	heta \sin 4	heta)$.$1 + 2 \cos 2	heta = 2(\cos 2	heta - \cos 6	heta)$.$1 + 2 \cos 2	heta = 2 \cos 2	heta - 2 \cos 6	heta$.$1 = -2 \cos 6	heta \Rightarrow \cos 6	heta = -\frac{1}{2}$.$6	heta = 2n\pi \pm \frac{2\pi}{3} \Rightarrow 	heta = \frac{n\pi}{3} \pm \frac{\pi}{9}$.For $n=0, 	heta = \frac{\pi}{9}$.For $n=1, 	heta = \frac{2\pi}{9}, \frac{4\pi}{9}$.For $n=2, 	heta = \frac{5\pi}{9}, \frac{7\pi}{9}$.For $n=3, 	heta = \frac{8\pi}{9}$.Total solutions: $0, \pi, \frac{\pi}{9}, \frac{2\pi}{9}, \frac{4\pi}{9}, \frac{5\pi}{9}, \frac{7\pi}{9}, \frac{8\pi}{9}$,which is $8$ solutions.

The equation $\sin 3\theta = 4 \sin \theta \sin 2\theta \sin 4\theta$ in the interval $0 \le \theta \le \pi$ has:

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