\[\begin{array}{l}
y = 1 + \sin 2x - {\cos ^2}2x = 1 + \sin 2x - \left( {1 - {{\sin }^2}2x} \right)\\
= {\sin ^2}2x + \sin 2x = {\sin ^2}2x + 2.\frac{1}{2}\sin 2x + \frac{1}{4} - \frac{1}{4}\\
= {\left( {\sin 2x + \frac{1}{2}} \right)^2} - \frac{1}{4} \ge - \frac{1}{4}\\
Dau\, = \,xay\,ra\,khi\,\sin 2x = - \frac{1}{2} \Rightarrow \min y = - \frac{1}{4}
\end{array}\]