Đáp án:
$\begin{array}{l}
2{\sin ^2}\left( {x + \frac{\pi }{4}} \right) + \cos 4x = 0\\
\Rightarrow {\left( {\sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right)} \right)^2} + \cos 4x = 0\\
\Rightarrow {\left( {\sin x + \cos x} \right)^2} + \cos 4x = 0\\
\left( {do:\sin x + \cos x = \sqrt 2 \sin \left( {x + \frac{\pi }{4}} \right)} \right)\\
\Rightarrow {\sin ^2}x + {\cos ^2}x + 2\sin x.\cos x + \cos 4x = 0\\
\Rightarrow 1 + \sin 2x + \left( {1 - 2{{\sin }^2}2x} \right) = 0\\
\Rightarrow - 2{\sin ^2}2x + \sin 2x + 2 = 0\\
\Rightarrow \sin 2x = \frac{{1 - \sqrt {17} }}{4}\left( {do:\sin 2x \le 1\forall x} \right)\\
\Rightarrow \left[ \begin{array}{l}
2x = \arcsin \frac{{1 - \sqrt {17} }}{4} + k2\pi \\
2x = \pi - \arcsin \frac{{1 - \sqrt {17} }}{4} + k2\pi
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \frac{{\arcsin \frac{{1 - \sqrt {17} }}{4}}}{2} + k\pi \\
x = \frac{{\pi - \arcsin \frac{{1 - \sqrt {17} }}{4}}}{2} + k\pi
\end{array} \right.
\end{array}$
