Let $S=\left\{\sin ^2 2 heta:\left(\sin ^4 heta+\cos ^4 heta ight) x^2+(\sin 2 heta) x+\left(\sin ^6 heta+\cos ^6 heta ight)=0 ight.$ has real roots $\}$. If $\alpha$ and $\beta$ be the smallest and largest elements of the set $S$, respectively, then…

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Let $S=\left\{\sin ^2 2 	heta:\left(\sin ^4 	heta+\cos ^4 	hetaight) x^2+(\sin 2 	heta) x+\left(\sin ^6 	heta+\cos ^6 	hetaight)=0ight.$ has real roots $\}$. If $\alpha$ and $\beta$ be the smallest and largest elements of the set $S$, respectively, then $3\left((\alpha-2)^2+(\beta-1)^2ight)$ equals _________

Aicharya · Accepted Answer

$\begin{aligned} & D=(\sin 2 	heta)^2-4\left(1-\frac{\sin ^2 2 	heta}{2}ight)\left(1-\frac{3}{4} \sin ^2 2 	hetaight) \ & =(\sin 2 	heta)^2-4\left(1-\frac{5}{4} \sin ^2 2 	heta+\frac{3}{8} \sin ^4 2 	hetaight) \ & D=-\frac{3}{2} \sin ^4 2 	heta+6 \sin ^2 2 	heta-4>0 \ & 3 \sin ^4 2 	heta-12 \sin ^2 2 	heta+8 < 0 \ & \sin ^2 2 	heta=\frac{12 \pm \sqrt{12^2-12.8}}{6}=\frac{12 \pm 4 \sqrt{3}}{6}=\frac{6 \pm 2 \sqrt{3}}{3} \ & \sin ^2 2 	heta=2 \pm \frac{2}{\sqrt{3}}, 	ext { but } \sin ^2 2 	heta \in[0,1] \ & 	herefore \alpha=2-\frac{2}{\sqrt{3}}, \beta=1 ightarrow(\alpha-2)^2=\frac{4}{3},(\beta-1)^2=0 \ & 3(\alpha-2)^2+(\beta-1)^2=4\end{aligned}$

Let $S=\left\{\sin ^2 2 \theta:\left(\sin ^4 \theta+\cos ^4 \theta\right) x^2+(\sin 2 \theta) x+\left(\sin…

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