\(y=2sin^2x+3sinx.cosx+cos^2x\)
\(=-\left(1-2sin^2x\right)+\dfrac{3}{2}sin2x+\dfrac{1}{2}\left(2cos^2x-1\right)+\dfrac{1}{2}\)
\(=-cos2x+\dfrac{3}{2}sin2x+\dfrac{1}{2}cos2x+\dfrac{1}{2}\)
\(=\dfrac{3}{2}sin2x-\dfrac{1}{2}cos2x+\dfrac{1}{2}\)
\(=\dfrac{\sqrt{10}}{2}\left(\dfrac{3}{\sqrt{10}}sin2x-\dfrac{1}{\sqrt{10}}cos2x\right)+\dfrac{1}{2}\)
\(=\dfrac{\sqrt{10}}{2}sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)+\dfrac{1}{2}\)
Vì \(sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)\in\left[-1;1\right]\)
\(\Rightarrow y=\dfrac{\sqrt{10}}{2}sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)+\dfrac{1}{2}\in\left[-\dfrac{\sqrt{10}}{2}+\dfrac{1}{2};\dfrac{\sqrt{10}}{2}+\dfrac{1}{2}\right]\)
\(\Rightarrow y_{min}=-\dfrac{\sqrt{10}}{2}+\dfrac{1}{2}\Leftrightarrow sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)=-1\Leftrightarrow...\)
\(y_{max}=\dfrac{\sqrt{10}}{2}+\dfrac{1}{2}\Leftrightarrow sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)=1\Leftrightarrow...\)
