\(y=cos2x-\sqrt{3}sin2x+2=2\left(\dfrac{1}{2}cos2x-\dfrac{\sqrt{3}}{2}sin2x\right)+2\)
\(=2coss\left(2x+\dfrac{\pi}{3}\right)+2\)
Do \(-1\le cos\left(2x+\dfrac{\pi}{3}\right)\le1\Rightarrow0\le y\le4\)
\(\Rightarrow\) C là đáp án đúng do \(\left(0;4\right)\subset\left[0;4\right]\)