Giải pt sau : $\cos x+\cos 3 x=1+\sqrt{2} \sin \left(2 x+\frac{\pi}{4}\right)$

$\begin{array}{l}
\cos x + \cos 3x = 1 + \sqrt 2 \sin \left( {2x + \dfrac{\pi }{4}} \right)\\
 \Leftrightarrow \cos x + \cos 3x = 1 + \sin 2x + \cos 2x\\
 \Leftrightarrow 2\cos 2x\cos x = 1 + 2\sin x\cos x + 2{\cos ^2}x - 1\\
 \Leftrightarrow 2\cos 2x\cos x = 2{\cos ^2}x + 2\sin x\cos x\\
 \Leftrightarrow \cos 2x\cos x = {\cos ^2}x + \sin x\cos x\\
 \Leftrightarrow \cos x\left( {\cos 2x - \cos x - \sin x} \right) = 0\\
 \Leftrightarrow \cos x\left[ {\left( {\cos x - \sin x} \right)\left( {\cos x + \sin x} \right) - \left( {\cos x + \sin x} \right)} \right] = 0\\
 \Leftrightarrow \cos x\left( {\cos x - \sin x} \right)\left( {\cos x + \sin x - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\cos x - \sin x = 0\\
\cos x + \sin x = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
\sqrt 2 \cos \left( {x + \dfrac{\pi }{4}} \right) = 0\\
\sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
\cos \left( {x + \dfrac{\pi }{4}} \right) = 0\\
\sin \left( {x + \dfrac{\pi }{4}} \right) = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x + \dfrac{\pi }{4} = \dfrac{\pi }{2} + k\pi \\
x + \dfrac{\pi }{4} = k\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = \dfrac{\pi }{4} + k\pi \\
x =  - \dfrac{\pi }{4} + k\pi 
\end{array} \right.\left( {k \in \mathbb Z} \right)
\end{array}$