The number of distinct solutions of the equat | vedclass

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(D) Given equation: $\frac{5}{4} \cos ^2 2x + (\cos ^4 x + \sin ^4 x) + (\cos ^6 x + \sin ^6 x) = 2$We know that $\cos ^4 x + \sin ^4 x = 1 - 2 \sin ^

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$8$

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(D) Given equation: $\frac{5}{4} \cos ^2 2x + (\cos ^4 x + \sin ^4 x) + (\cos ^6 x + \sin ^6 x) = 2$We know that $\cos ^4 x + \sin ^4 x = 1 - 2 \sin ^2 x \cos ^2 x = 1 - \frac{1}{2} \sin ^2 2x$And $\cos ^6 x + \sin ^6 x = 1 - 3 \sin ^2 x \cos ^2 x = 1 - \frac{3}{4} \sin ^2 2x$Substituting these into the equation:$\frac{5}{4} \cos ^2 2x + (1 - \frac{1}{2} \sin ^2 2x) + (1 - \frac{3}{4} \sin ^2 2x) = 2$$\frac{5}{4} \cos ^2 2x + 2 - \frac{5}{4} \sin ^2 2x = 2$$\frac{5}{4} (\cos ^2 2x - \sin ^2 2x) = 0$$\frac{5}{4} \cos 4x = 0$$\cos 4x = 0$,where $x \in [0, 2\pi] \Rightarrow 4x \in [0, 8\pi]$The values of $4x$ for which $\cos 4x = 0$ are $\frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}, \frac{9\pi}{2}, \frac{11\pi}{2}, \frac{13\pi}{2}, \frac{15\pi}{2}$Thus,there are $8$ distinct solutions.

The number of distinct solutions of the equation $\frac{5}{4} \cos ^2 2x + \cos ^4 x + \sin ^4 x + \cos ^6 x + \sin ^6 x = 2$ in the interval $[0, 2\pi]$ is

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