If theta in [-2 pi, 2 pi],then the number of | vedclass

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(C) Given equation: $2 \sqrt{2} \cos^2 	heta + (2 - \sqrt{6}) \cos 	heta - \sqrt{3} = 0$ Factorizing the quadratic equation: $2 \sqrt{2} \cos^2 	he

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$8$

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(C) Given equation: $2 \sqrt{2} \cos^2 	heta + (2 - \sqrt{6}) \cos 	heta - \sqrt{3} = 0$ Factorizing the quadratic equation: $2 \sqrt{2} \cos^2 	heta + 2 \cos 	heta - \sqrt{6} \cos 	heta - \sqrt{3} = 0$ $2 \cos 	heta (\sqrt{2} \cos 	heta + 1) - \sqrt{3} (\sqrt{2} \cos 	heta + 1) = 0$ $(2 \cos 	heta - \sqrt{3})(\sqrt{2} \cos 	heta + 1) = 0$ This gives two cases: $1) \cos 	heta = \frac{\sqrt{3}}{2}$ $2) \cos 	heta = -\frac{1}{\sqrt{2}}$ For $	heta \in [-2 \pi, 2 \pi]$: For $\cos 	heta = \frac{\sqrt{3}}{2}$,the solutions are $	heta = \pm \frac{\pi}{6}, \pm \frac{11 \pi}{6}$ ($4$ solutions). For $\cos 	heta = -\frac{1}{\sqrt{2}}$,the solutions are $	heta = \pm \frac{3 \pi}{4}, \pm \frac{5 \pi}{4}$ ($4$ solutions). Total number of solutions $= 4 + 4 = 8$.

If $\theta \in [-2 \pi, 2 \pi]$,then the number of solutions of $2 \sqrt{2} \cos^2 \theta + (2 - \sqrt{6}) \cos \theta - \sqrt{3} = 0$ is equal to:

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