Let S= sin ^2 2 θ: (sin ^4 θ+cos ^4 θ ) x^2+(sin 2 θ) x+ (sin ^6 θ+cos ^6 θ )=0 . has real roots . If α and β be the smallest and largest elements of…

$$\begin{aligned} & D=(\sin 2 \theta)^2-4\left(1-\frac{\sin ^2 2 \theta}{2}\right)\left(1-\frac{3}{4} \sin ^2 2 \theta\right) \\ & =(\sin 2 \theta)^2-4\left(1-\frac{5}{4} \sin ^2 2 \theta+\frac{3}{8} \sin ^4 2 \theta\right) \\ & D=-\frac{3}{2} \sin ^4 2 \theta+6 \sin ^2 2 \theta-4>0 \\ & 3 \sin ^4 2 \theta-12 \sin ^2 2 \theta+8 < 0 \\ & \sin ^2 2 \theta=\frac{12 \pm \sqrt{12^2-12.8}}{6}=\frac{12 \pm 4 \sqrt{3}}{6}=\frac{6 \pm 2 \sqrt{3}}{3} \\ & \sin ^2 2 \theta=2 \pm \frac{2}{\sqrt{3}}, \text { but } \sin ^2 2 \theta \in[0,1] \\ & \therefore \alpha=2-\frac{2}{\sqrt{3}}, \beta=1 \rightarrow(\alpha-2)^2=\frac{4}{3},(\beta-1)^2=0 \\ & 3(\alpha-2)^2+(\beta-1)^2=4\end{aligned}$$​D=(sin2θ)2−4(1−2sin22θ​)(1−43​sin22θ)=(sin2θ)2−4(1−45​sin22θ+83​sin42θ)D=−23​sin42θ+6sin22θ−4>03sin42θ−12sin22θ+8<0sin22θ=612±122−12.8​​=612±43​​=36±23​​sin22θ=2±3​2​, but sin22θ∈[0,1]∴α=2−3​2​,β=1→(α−2)2=34​,(β−1)2=03(α−2)2+(β−1)2=4​