`~rai~`
\(y=2\sin x+2\cos x-2\sin2x\\\quad=(2\sin x+\cos x)-2.2\sin x\cos x\\\quad=2(\sin x+\cos x)-4\sin x\cos x.(1)\\\text{Đặt }t=\sin x+\cos x\quad(|t|\le \sqrt{2})\\\Leftrightarrow t^2=(\sin x+\cos x)^2\\\Leftrightarrow t^2=\sin^2x+2\sin x\cos x+\cos^2x\\\Leftrightarrow t^2=1+2\sin x\cos x\\\Leftrightarrow 2\sin x\cos x=t^2-1.\\\text{Thay vào (1) được:}\\y=2t-2(t^2-1)\\\quad=-2t^2+2t+2.\\\text{Bảng biến thiên:}\\\begin{array}{|c|cr|}\hline t&-\sqrt{2}&&&&&\dfrac{1}{2}&&&&&\sqrt{2}\\\hline&&&&\nearrow&&\dfrac{5}{2}&&\searrow\\y&-2-\sqrt{2}&&&&&&&&&&-2+\sqrt{2}\\\hline\end{array}\\\text{Vậy Min}_y=-2-\sqrt{2};\text{Max}_y=\dfrac{5}{2}.\)
