Let S=left{sin ^2 2 theta:left(sin ^4 theta+c | vedclass

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Question

$ \mathrm{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(

Vedclass · Accepted Answer

$4$

Vedclass · Answer

$ \mathrm{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) $

$ \mathrm{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$

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