The number of values of x satisfying 2sin^2(2 | vedclass

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(A) Given equation: $2\sin^2(2x) = 2\cos^2(8x) + \cos(10x)$Using $2\sin^2(	heta) = 1 - \cos(2	heta)$ and $2\cos^2(	heta) = 1 + \cos(2	heta)$:$1 -

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$10$

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(A) Given equation: $2\sin^2(2x) = 2\cos^2(8x) + \cos(10x)$Using $2\sin^2(	heta) = 1 - \cos(2	heta)$ and $2\cos^2(	heta) = 1 + \cos(2	heta)$:$1 - \cos(4x) = 1 + \cos(16x) + \cos(10x)$$\cos(16x) + \cos(10x) + \cos(4x) = 0$Using $\cos(16x) + \cos(4x) = 2\cos(10x)\cos(6x)$:$2\cos(10x)\cos(6x) + \cos(10x) = 0$$\cos(10x)(2\cos(6x) + 1) = 0$Case $1$: $\cos(10x) = 0 \implies 10x = (2n+1)\frac{\pi}{2} \implies x = (2n+1)\frac{\pi}{20}$.For $x \in [-\frac{\pi}{4}, \frac{\pi}{4}]$,$10x \in [-\frac{5\pi}{2}, \frac{5\pi}{2}]$. Values are $n = -2, -1, 0, 1, 2$ (Total $5$ values).Case $2$: $\cos(6x) = -\frac{1}{2} \implies 6x = 2n\pi \pm \frac{2\pi}{3} \implies x = \frac{n\pi}{3} \pm \frac{\pi}{9}$.For $x \in [-\frac{\pi}{4}, \frac{\pi}{4}]$,$6x \in [-\frac{3\pi}{2}, \frac{3\pi}{2}]$.Solutions: $x = \pm \frac{\pi}{9}, \pm \frac{2\pi}{9}, \pm \frac{4\pi}{9}$ (only $\pm \frac{\pi}{9}, \pm \frac{2\pi}{9}$ are in range,total $4$ values).Wait,checking range: $x = \frac{n\pi}{3} \pm \frac{\pi}{9}$. For $n=0: \pm \frac{\pi}{9}$. For $n=1: \frac{3\pi \pm \pi}{9} = \frac{4\pi}{9}, \frac{2\pi}{9}$. For $n=-1: \frac{-3\pi \pm \pi}{9} = -\frac{4\pi}{9}, -\frac{2\pi}{9}$.Values in $[-\frac{\pi}{4}, \frac{\pi}{4}]$ are $\pm \frac{\pi}{9}, \pm \frac{2\pi}{9}$. Total $4$ values.Total values $= 5 + 4 = 9$. Re-evaluating: $10x = \pm \frac{\pi}{2}, \pm \frac{3\pi}{2}, \pm \frac{5\pi}{2} \implies x = \pm \frac{\pi}{20}, \pm \frac{3\pi}{20}, \pm \frac{5\pi}{20} = \pm \frac{\pi}{4}$. Total $6$ values.Total $= 6 + 4 = 10$.

The number of values of $x$ satisfying $2\sin^2(2x) = 2\cos^2(8x) + \cos(10x)$ in the interval $x \in \left[ -\frac{\pi}{4}, \frac{\pi}{4} \right]$ is:

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