\(y=4cos^2\left(\dfrac{x}{2}-\dfrac{\pi}{12}\right)-7=2\left[cos\left(x-\dfrac{\pi}{6}\right)+1\right]-7=2cos\left(x-\dfrac{\pi}{6}\right)-5\)
Đặt \(x-\dfrac{\pi}{6}=t\Rightarrow t\in\left[-\dfrac{\pi}{6};\dfrac{5\pi}{6}\right]\)
\(\Rightarrow y=2cost-5\)
Do \(t\in\left[-\dfrac{\pi}{6};\dfrac{5\pi}{6}\right]\Rightarrow cost\in\left[-\dfrac{\sqrt{3}}{2};1\right]\)
\(\Rightarrow y\in\left[-5-\sqrt{3};-3\right]\)
\(y_{max}=-3\) khi \(t=0\) hay \(x=\dfrac{\pi}{6}\)
\(y_{min}=-5-\sqrt{3}\) khi \(y=\dfrac{5\pi}{6}\) hay \(x=\pi\)