$\begin{array}{l}
\cos x = - \dfrac{{\sqrt 2 }}{2}\\
\Leftrightarrow \cos x = \cos \left( {\dfrac{\pi }{2} + \dfrac{\pi }{4}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + \dfrac{\pi }{4} + k2\pi \\
x = - \dfrac{\pi }{2} - \dfrac{\pi }{4} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{3\pi }}{4} + k2\pi \\
x = - \dfrac{{3\pi }}{4} + k2\pi
\end{array} \right.\left( {k \in \mathbb{Z}} \right)\\
5.{\cos ^2}2x = \dfrac{1}{4}\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 2x = - \dfrac{1}{2}\\
\cos 2x = \dfrac{1}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} + \dfrac{\pi }{6} + k2\pi \\
2x = \dfrac{\pi }{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{3} + k\pi \\
x = \dfrac{\pi }{6} + k\pi
\end{array} \right.\left( {k \in \mathbb{Z}} \right)
\end{array}$
