\[\begin{array}{l}
y = 2sinxcosx + cos2x\\
= \sin 2x + \cos 2x\\
= \sqrt 2 .\left( {\frac{1}{{\sqrt 2 }}\sin 2x + \frac{1}{{\sqrt 2 }}\cos 2x} \right)\\
= \sqrt 2 .\left( {\sin 2x.\cos \frac{\pi }{4} + \cos 2x.\sin \frac{\pi }{4}} \right)\\
= \sqrt 2 .\sin \left( {2x + \frac{\pi }{4}} \right)\\
do: - 1 \le \sin \left( {2x + \frac{\pi }{4}} \right) \le 1\\
= > - \sqrt 2 \le \sqrt 2 \sin \left( {2x + \frac{\pi }{4}} \right) \le \sqrt 2 \\
= > - \sqrt 2 \le y \le \sqrt 2
\end{array}\]

\[\begin{array}{l}
+ )y = - \sqrt 2 \Leftrightarrow \sin \left( {2x + \frac{\pi }{4}} \right) = - 1 \Leftrightarrow 2x + \frac{\pi }{4} = \frac{{3\pi }}{2} + k2\pi \Leftrightarrow x = \frac{{5\pi }}{8} + k\pi \left( {k \in Z} \right)\\
+ )y = \sqrt 2 \Leftrightarrow \sin \left( {2x + \frac{\pi }{4}} \right) = 1 \Leftrightarrow 2x + \frac{\pi }{4} = \frac{\pi }{2} + k2\pi \Leftrightarrow x = \frac{\pi }{8} + k\pi \left( {k \in Z} \right)
\end{array}\]

vậy y max=\[\sqrt 2 \Leftrightarrow x = \frac{\pi }{8} + k\pi \left( {k \in Z} \right)\]