\(0\leq cos^2\left(2x-\dfrac{\pi}{3} \right) \leq 1\Rightarrow 0 \leq 2cos^2\left(2x-\dfrac{\pi}{3} \right) \leq 2\)
\(\Rightarrow -5 \leq -5+2cos^2\left(2x-\dfrac{\pi}{3} \right) \leq -3\)
Vậy \(y_{min}=-5\) khi \(cos^2\left(2x-\dfrac{\pi}{3} \right)=0\)
\(y_{max}=-3\) khi \(cos^2\left(2x-\dfrac{\pi}{3} \right)=1\)