If 4(sin 2x sin 4x + sin^2 x) = 3,then x = | vedclass

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(A) Given the equation: $4(\sin 2x \sin 4x + \sin^2 x) = 3$ Using the identity $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$ and $2 \sin^2 x = 1 - \cos 2x

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$\frac{n \pi}{3} \pm \frac{\pi}{9}, n \in Z$

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(A) Given the equation: $4(\sin 2x \sin 4x + \sin^2 x) = 3$ Using the identity $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$ and $2 \sin^2 x = 1 - \cos 2x$: $2(2 \sin 2x \sin 4x) + 2(2 \sin^2 x) = 3$ $2(\cos(4x-2x) - \cos(4x+2x)) + 2(1 - \cos 2x) = 3$ $2(\cos 2x - \cos 6x) + 2 - 2 \cos 2x = 3$ $2 \cos 2x - 2 \cos 6x + 2 - 2 \cos 2x = 3$ $-2 \cos 6x + 2 = 3$ $-2 \cos 6x = 1$ $\cos 6x = -\frac{1}{2}$ Since $\cos 	heta = \cos \alpha \Rightarrow 	heta = 2n\pi \pm \alpha$,where $\alpha = \frac{2\pi}{3}$: $6x = 2n\pi \pm \frac{2\pi}{3}$ Dividing by $6$: $x = \frac{n\pi}{3} \pm \frac{\pi}{9}, n \in Z$

If $4(\sin 2x \sin 4x + \sin^2 x) = 3$,then $x =$

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