$ \mathrm{D}=(\sin 2 \theta)^2-4\left(1-\frac{\sin ^2 2 \theta}{2}\right)\left(1-\frac{3}{4} \sin ^2 2 \theta\right) $

$ =(\sin 2 \theta)^2-4\left(1-\frac{5}{4} \sin ^2 2 \theta+\frac{3}{8} \sin ^4 2 \theta\right) $

$ \mathrm{D}=-\frac{3}{2} \sin ^4 2 \theta+6 \sin ^2 2 \theta-4>0 $

$ 3 \sin ^4 2 \theta-12 \sin ^2 2 \theta+8<0 $

$ \sin ^2 2 \theta=\frac{12 \pm \sqrt{12^2-12.8}}{6}=\frac{12 \pm 4 \sqrt{3}}{6}=\frac{6 \pm 2 \sqrt{3}}{3} $

$ \sin ^2 2 \theta=2 \pm \frac{2}{\sqrt{3}}, \text { but } \sin ^2 2 \theta \in[0,1] $

$ \therefore \alpha=2-\frac{2}{\sqrt{3}}, \beta=1 \rightarrow(\alpha-2)^2=\frac{4}{3},(\beta-1)^2=0 $

$ 3(\alpha-2)^2+(\beta-1)^2=4$