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

Question

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

Aicharya · Accepted Answer

$\begin{aligned} & 2 \sqrt{2} \cos ^2 	heta+2 \cos 	heta-\sqrt{6} \cos 	heta-\sqrt{3}=0 \ & (2 \cos 	heta-\sqrt{3})(\sqrt{2} \cos 	heta+1)=0 \ & \cos 	heta=\frac{\sqrt{3}}{2}, \frac{-1}{\sqrt{2}}\end{aligned}$ Number of solution $=8$ *

If $\theta \in[-2 \pi, 2 \pi]$, then the number of solutions of $2 \sqrt{2} \cos ^2 \theta+(2-\sqrt{6}) \cos…

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