GTLN hàm số y=2.cosx^2+x với 0<=x<=pi/2

$y=2{\cos}^2x+x=1+\cos 2x+x$ 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 $x=\dfrac{5\pi}{12}$