Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
y = \sin \left( {2x + \frac{\pi }{4}} \right)\\
 \Rightarrow y' = \left( {2x + \frac{\pi }{4}} \right)'.cos\left( {2x + \frac{\pi }{4}} \right) = 2\cos \left( {2x + \frac{\pi }{4}} \right)\\
y' = 0 \Leftrightarrow \cos \left( {2x + \frac{\pi }{4}} \right) = 0 \Leftrightarrow 2x + \frac{\pi }{4} = \frac{\pi }{2} + k\pi  \Leftrightarrow x = \frac{\pi }{8} + \frac{{k\pi }}{2}\\
b,\\
y = \cos 2x + x\\
 \Rightarrow y' = \left( {2x} \right)'.\left( { - \sin 2x} \right) + 1 =  - 2\sin 2x + 1\\
y' = 0 \Leftrightarrow \sin 2x = \frac{1}{2} \Leftrightarrow \left[ \begin{array}{l}
2x = \frac{\pi }{6} + k2\pi \\
2x = \frac{{5\pi }}{6} + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{{12}} + k\pi \\
x = \frac{{5\pi }}{{12}} + k\pi 
\end{array} \right.\\
c,\\
y = \sin x + \cos x + \frac{x}{{\sqrt 2 }}\\
 \Rightarrow y' = \cos x - \sin x + \frac{1}{{\sqrt 2 }}\\
y' = 0 \Leftrightarrow \sin x - \cos x = \frac{1}{{\sqrt 2 }}\\
 \Leftrightarrow \sqrt 2 .\sin \left( {x - \frac{\pi }{4}} \right) = \frac{1}{{\sqrt 2 }}\\
 \Leftrightarrow \sin \left( {x - \frac{\pi }{4}} \right) = \frac{1}{2}\\
 \Leftrightarrow \left[ \begin{array}{l}
x - \frac{\pi }{4} = \frac{\pi }{6} + k2\pi \\
x - \frac{\pi }{4} = \frac{{5\pi }}{6} + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{{5\pi }}{{12}} + k2\pi \\
x = \frac{{13\pi }}{{12}} + k2\pi 
\end{array} \right.\\
d,\\
y = \sqrt 3 \sin x + \cos x + 2x\\
 \Rightarrow y' = \sqrt 3 \cos x - \sin x + 2\\
y' = 0 \Leftrightarrow \sin x - \sqrt 3 \cos x = 2\\
 \Leftrightarrow \frac{1}{2}\sin x - \frac{{\sqrt 3 }}{2}\cos x = 1\\
 \Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = 1\\
 \Leftrightarrow x - \frac{\pi }{3} = \frac{\pi }{2} + k2\pi \\
 \Leftrightarrow x = \frac{{5\pi }}{6} + k2\pi \\
e,\\
y = {\sin ^2}2x - 2{\cos ^2}x\\
 \Rightarrow y' = 2.\left( {\sin 2x} \right)'.\sin 2x - 2.2.\left( {\cos x} \right)'.\cos x\\
 = 2.2.\cos 2x.\sin 2x - 2.2.\left( { - \sin x} \right).\cos x\\
 = 2.sin4x + 2.\sin 2x\\
y' = 0 \Leftrightarrow \sin 4x =  - \sin 2x \Leftrightarrow \sin 4x = \sin \left( { - 2x} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
4x =  - 2x + k2\pi \\
4x = \pi  + 2x + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{{k\pi }}{3}\\
x = \frac{\pi }{2} + k\pi 
\end{array} \right.
\end{array}\)