Ta có: \(sin^2x\le1\Rightarrow y\ge2sinx+\sqrt{3-1}=2sinx+\sqrt{2}\)
Mặt khác \(sinx\ge-1\Rightarrow y\ge-2+\sqrt{2}\)
\(\Rightarrow y_{min}=-2+\sqrt{2}\) khi \(sinx=-1\)
\(y^2=3sin^2x+3+4sinx\sqrt{3-sin^2x}\)
\(y^2=3sin^2x+3+2\sqrt{2}\left(\sqrt{2}sinx.\sqrt{3-sin^2x}\right)\)
\(y^2\le3sin^2x+3+\sqrt{2}\left(2sin^2x+3-sin^2x\right)=\left(3+\sqrt{2}\right)sin^2x+3+3\sqrt{2}\)
Do \(sin^2x\le1\Rightarrow y^2\le3+\sqrt{2}+3+3\sqrt{2}=6+4\sqrt{2}\)
\(\Rightarrow y\le\sqrt{6+4\sqrt{2}}=2+\sqrt{2}\)
\(y_{max}=2+\sqrt{2}\) khi \(sinx=1\)
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