Tìm GTLN – GTNN của y = (2sinx + cosx)(3sinx – cosx)

$\begin{array}{l} y = \left( {2\sin x + \cos x} \right)\left( {3\sin x – \cos x} \right)\\ y = 6{\sin ^2}x – 2\sin x\cos x + 3\sin x\cos x – {\cos ^2}x\\ y = 7{\sin ^2}x + \sin x\cos x – \left( {{{\sin }^2}x + {{\cos }^2}x} \right)\\ y = 7{\sin ^2}x + \sin x\cos x – 1\\ y = \dfrac{{7\left( {1 – \cos 2x} \right)}}{2} + \dfrac{1}{2}\sin 2x – 1\\ y = \dfrac{{ – 7}}{2}\cos 2x + \dfrac{1}{2}\sin 2x + \dfrac{5}{2}\\ y = \dfrac{1}{2}\left( {\sin 2x – 7\cos 2x} \right) + \dfrac{5}{2}\\ y = \dfrac{1}{2}.\sqrt {50} \left( {\dfrac{1}{{\sqrt {50} }}\sin 2x – \dfrac{7}{{\sqrt {50} }}\cos 2x} \right) + \dfrac{5}{2}\\ y = \dfrac{{\sqrt {50} }}{2}\sin \left( {x – \alpha } \right) + 3\left( {\alpha  = \arccos \dfrac{1}{{\sqrt {50} }}} \right)\\  – 1 \le \sin \left( {x – \alpha } \right) \le 1\\  \Rightarrow \dfrac{5}{2} – \dfrac{{\sqrt {50} }}{2} \le y \le \dfrac{5}{2} + \dfrac{{\sqrt {50} }}{2}\\  \Rightarrow \left\{ \begin{array}{l} \max y = \dfrac{5}{2} + \dfrac{{\sqrt {50} }}{2}\\ \min y = \dfrac{5}{2} – \dfrac{{\sqrt {50} }}{2} \end{array} \right. \end{array}$