$y=2{\cos}^2x+x=1+\cos 2x+x$
TXĐ: $D=\mathbb R$
$y'=-2\sin 2x+1=0$
$\Rightarrow\sin 2x=\dfrac{1}{2}$
$\Rightarrow\left[\begin{array}{l} 2x=\dfrac{\pi}{6} \\ 2x=\dfrac{5\pi}{6}\end{array} \right .$
Do $x\in[0;\dfrac{\pi}{2}]\Rightarrow 2x\in[0;\pi]$
$\Rightarrow \left[ \begin{array}{l} x=\dfrac{\pi}{12} \\ x=\dfrac{5\pi}{12}\end{array} \right .$
Ta có: $y(0)=1+\cos (2.0)+0=2$
$y(\dfrac{\pi}{12})=1+\cos (2.\dfrac{\pi}{12})+\dfrac{\pi}{12}=1+\dfrac{\sqrt3}{2}+\dfrac{\pi}{12}$
$y(\dfrac{5\pi}{12})=1+\cos (2.\dfrac{5\pi}{12})+\dfrac{5\pi}{12}=1-\dfrac{\sqrt3}{2}+\dfrac{5\pi}{12}$
$y(\dfrac{\pi}{2})=1+\cos (2.\dfrac{\pi}{2})+\dfrac{\pi}{2}=1-1+\dfrac{\pi}{2}=\dfrac{\pi}{2}$
Suy ra GTLN là $y=1+\dfrac{\sqrt3}{2}+\dfrac{\pi}{12}$ tại $x=\dfrac{\pi}{12}$
GTNN là $y=1-\dfrac{\sqrt3}{2}+\dfrac{5\pi}{12}$ tai $x=\dfrac{5\pi}{12}$
