\(y=1-cos2x+2sin2x+6=2sin2x-cos2x+7\)
\(y=\sqrt{5}\left(\dfrac{2}{\sqrt{5}}sin2x-\dfrac{1}{\sqrt{5}}cos2x\right)+7\)
Đặt \(\dfrac{2}{\sqrt{5}}=cosa\) với \(a\in\left(0;\dfrac{\pi}{2}\right)\)
\(y=\sqrt{5}sin\left(2x-a\right)+7\)
\(\Rightarrow-\sqrt{5}+7\le y\le\sqrt{5}+7\)