If f(x)=cos^2 x+cos^2 2x+cos^2 3x,then the nu | vedclass

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(D) Given $f(x) = \cos^2 x + \cos^2 2x + \cos^2 3x = 1$. Using the identity $\cos^2 	heta = \frac{1+\cos 2	heta}{2}$,we have: $f(x) = \frac{1+\cos 2

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$10$

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(D) Given $f(x) = \cos^2 x + \cos^2 2x + \cos^2 3x = 1$. Using the identity $\cos^2 	heta = \frac{1+\cos 2	heta}{2}$,we have: $f(x) = \frac{1+\cos 2x}{2} + \frac{1+\cos 4x}{2} + \cos^2 3x = 1$ $1 + \frac{1}{2}(\cos 2x + \cos 4x) + \cos^2 3x = 1$ $\frac{1}{2}(2\cos 3x \cos x) + \cos^2 3x = 0$ $\cos 3x (\cos x + \cos 3x) = 0$ $\cos 3x (2 \cos 2x \cos x) = 0$ $2 \cos 3x \cos 2x \cos x = 0$ This implies $\cos 3x = 0$ or $\cos 2x = 0$ or $\cos x = 0$. For $x \in [0, 2\pi]$: $1$. $\cos x = 0 \Rightarrow x = \frac{\pi}{2}, \frac{3\pi}{2}$ $2$. $\cos 2x = 0$ $\Rightarrow 2x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}$ $\Rightarrow x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$ $3$. $\cos 3x = 0$ $\Rightarrow 3x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}, \frac{9\pi}{2}, \frac{11\pi}{2}$ $\Rightarrow x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}$ Combining all unique values: $x \in \{\frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{5\pi}{4}, \frac{3\pi}{2}, \frac{7\pi}{4}, \frac{11\pi}{6}\}$. Counting these,there are $10$ distinct values.

If $f(x)=\cos^2 x+\cos^2 2x+\cos^2 3x$,then the number of values of $x \in [0, 2\pi]$ for which $f(x)=1$ is

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