a) Đặt \(t=sinx+cosx\)
\(\Rightarrow t^2=\left(sinx+cosx\right)^2\overset{bunhiacopxki}{\le}\left(1^2+1^2\right)\left(sinx^2+cosx^2\right)=2\\ \Rightarrow-\sqrt{2}\le t\le\sqrt{2}\\ \Rightarrow-\sqrt{2}+1\le y=t+1\le\sqrt{2}+1\)
Vậy \(Min\text{ }y=-\sqrt{2}+1\Leftrightarrow sinx=cosx=\frac{-1}{\sqrt{2}}\Leftrightarrow x=\pm\frac{3\pi}{4}+k2\pi\)
\(Max\text{ }y=\sqrt{2}+1\Leftrightarrow sinx=cosx=\frac{1}{\sqrt{2}}\Leftrightarrow x=\pm\frac{\pi}{4}+k2\pi\)
\(b\text{) }y=cosx-cos2x+4\\ =cosx-\left(2cos^2x-1\right)+4\\ =-2cos^2x+cosx+5\)
\(\Rightarrow Min\text{ }y=2\Leftrightarrow cosx=-1\)
\(\Rightarrow Max\text{ }y=\frac{41}{8}\Leftrightarrow cosx=\frac{1}{4}\)
\(\text{c) }y=2sin^2x+4\sqrt{3}sinx\cdot cosx+6cos^2x+1\\ =\left(1-cos2x\right)+2\sqrt{3}sin2x+3\left(cos2x+1\right)+1\\ =2cos2x+2\sqrt{3}sin2x+5\)
Đặt \(t=2cos2x+2\sqrt{3}sin2x\)
\(\Rightarrow t^2\le\left[2^2+\left(2\sqrt{3}\right)^2\right]\left(cos^22x+sin^22x\right)=16\\ \Rightarrow-4\le t\le4\\ \Rightarrow1\le y\le9\\ \)
Vậy \(Min\text{ }y=1\Leftrightarrow sin2x=-\frac{1}{2}\)
\(Max\text{ }y=9\Leftrightarrow sin2x=\frac{1}{2}\)
