Lời giải:
Vì \(\cos ^2x; \sin ^2x\geq 0, \forall x\Rightarrow 5-2\cos^2x\sin ^2x\leq 5\)
\(\Rightarrow y=\sqrt{5-2\cos ^2x\sin ^2x}\leq \sqrt{5}\)
Vậy \(y_{\max}=\sqrt{5}\Leftrightarrow \sin x=0\) hoặc \(\cos x=0\)
\(y=\sqrt{5-2\cos ^2x\sin ^2x}=\sqrt{5-\frac{(2\sin x\cos x)^2}{2}}\)
\(=\sqrt{5-\frac{\sin ^22x}{2}}\)
Ta thấy: \(\sin ^22x\leq 1\Rightarrow 5-\frac{\sin ^22x}{2}\geq \frac{9}{2}\)
\(\Rightarrow y\geq \frac{3}{\sqrt{2}}\)
Vậy \(y_{\min}=\frac{3}{\sqrt{2}}\Leftrightarrow \sin 2x=\pm 1\)
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