$\begin{aligned} & 2 \sqrt{2} \cos ^2 \theta+(2-\sqrt{6}) \cos \theta-\sqrt{3}=0 \\ & 2 \sqrt{2} \cos ^2 \theta+2 \cos \theta-\sqrt{6} \cos \theta-\sqrt{3}=0 \\ & (2 \cos \theta-\sqrt{3})(\sqrt{2} \cos \theta+1)=0 \\ & \Rightarrow \cos \theta=\frac{\sqrt{3}}{2} \text { or } \cos \theta=\frac{-1}{\sqrt{2}} \\ & \theta=\left\{\frac{-11 \pi}{6}, \frac{-5 \pi}{4}, \frac{-3 \pi}{4}, \frac{-\pi}{6}, \frac{\pi}{6}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{11 \pi}{6}\right\} \\ & \Rightarrow 8 \text { (solution) } \end{aligned}$