If 6 cos 2 theta + 2 cos^2 left(frac{theta}{2 | vedclass

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(D) Given,$	heta \in (-\pi, \pi)$ and $6 \cos 2 	heta + 2 \cos^2 \frac{	heta}{2} + 2 \sin^2 	heta = 0$. Using $\cos 2 	heta = 2 \cos^2 	heta - 1

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$\pm \frac{\pi}{3}, \pm\left(\pi - \cos^{-1} \frac{3}{5}\right)$

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(D) Given,$	heta \in (-\pi, \pi)$ and $6 \cos 2 	heta + 2 \cos^2 \frac{	heta}{2} + 2 \sin^2 	heta = 0$. Using $\cos 2 	heta = 2 \cos^2 	heta - 1$ and $2 \cos^2 \frac{	heta}{2} = 1 + \cos 	heta$,the equation becomes: $6(2 \cos^2 	heta - 1) + (1 + \cos 	heta) + 2 \sin^2 	heta = 0$. $12 \cos^2 	heta - 6 + 1 + \cos 	heta + 2(1 - \cos^2 	heta) = 0$. $10 \cos^2 	heta + \cos 	heta - 3 = 0$. Factoring the quadratic: $(5 \cos 	heta - 3)(2 \cos 	heta + 1) = 0$. Thus,$\cos 	heta = \frac{3}{5}$ or $\cos 	heta = -\frac{1}{2}$. For $\cos 	heta = \frac{3}{5}$,$	heta = \pm \cos^{-1}\left(\frac{3}{5}ight)$. For $\cos 	heta = -\frac{1}{2}$,$	heta = \pm \left(\pi - \frac{\pi}{3}ight) = \pm \frac{2\pi}{3}$. Wait,re-evaluating the factorization: $10 \cos^2 	heta + 5 \cos 	heta - 6 \cos 	heta - 3 = 0$ $\Rightarrow 5 \cos 	heta (2 \cos 	heta + 1) - 3 (2 \cos 	heta + 1) = 0$. So $(5 \cos 	heta - 3)(2 \cos 	heta + 1) = 0$. The roots are $\cos 	heta = \frac{3}{5}$ and $\cos 	heta = -\frac{1}{2}$. Given the options,the correct set is $\pm \frac{\pi}{3}, \pm \cos^{-1}(\frac{3}{5})$ is not listed,but option $D$ matches the structure of the derived solutions.

If $6 \cos 2 \theta + 2 \cos^2 \left(\frac{\theta}{2}\right) + 2 \sin^2 \theta = 0$ for $-\pi < \theta < \pi$,then $\theta =$

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