$3-\dfrac{1}{2}\cos2x$
Ta có: $-1\le\cos x\le1$ $\forall x$
$\Rightarrow -1\le\cos2x\le1$
$\Rightarrow -\dfrac{1}{2}\le\dfrac{1}{2}\cos2x\le\dfrac{1}{2}$
$\Rightarrow \dfrac{1}{2}\ge-\dfrac{1}{2}\cos2x\ge-\dfrac{1}{2}$
$\Rightarrow 3+\dfrac{1}{2}\ge3-\dfrac{1}{2}\cos2x\ge3-\dfrac{1}{2}$
$\Rightarrow \dfrac{7}{2}\ge3-\dfrac{1}{2}\cos2x\ge\dfrac{5}{2}$
Vậy hàm số đạt GTNN bằng $\dfrac{5}{2}$ tại $\cos 2x=1$;
hàm số đạt GTLN bằng $\dfrac{7}{2}$ tại $\cos 2x=-1$.
