Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
e,\\
2{\cos ^2}2x = 1\\
 \Leftrightarrow {\cos ^2}2x = \dfrac{1}{2}\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos 2x = \dfrac{{\sqrt 2 }}{2}\\
\cos 2x =  - \dfrac{{\sqrt 2 }}{2}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x =  \pm \dfrac{\pi }{4} + k2\pi \\
2x =  \pm \dfrac{{3\pi }}{4} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow 2x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\
 \Leftrightarrow x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{4}\\
f,\\
4{\cos ^2}4x = 1\\
 \Leftrightarrow {\cos ^2}4x = \dfrac{1}{4}\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos 4x = \dfrac{1}{2}\\
\cos 4x =  - \dfrac{1}{2}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
4x =  \pm \dfrac{\pi }{3} + k2\pi \\
4x =  \pm \dfrac{{2\pi }}{3} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  \pm \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{2}\\
x =  \pm \dfrac{\pi }{6} + \dfrac{{k\pi }}{2}
\end{array} \right.\\
g,\\
4{\cos ^2}5x = 3\\
 \Leftrightarrow {\cos ^2}5x = \dfrac{3}{4}\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos 5x = \dfrac{{\sqrt 3 }}{2}\\
\cos 5x =  - \dfrac{{\sqrt 3 }}{2}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
5x =  \pm \dfrac{\pi }{6} + k2\pi \\
5x =  \pm \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  \pm \dfrac{\pi }{{30}} + \dfrac{{k2\pi }}{5}\\
x =  \pm \dfrac{\pi }{6} + \dfrac{{k2\pi }}{5}
\end{array} \right.\\
h,\\
8{\cos ^2}x = 3\\
 \Leftrightarrow {\cos ^2}x = \dfrac{3}{8}\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x = \dfrac{{\sqrt 6 }}{4}\\
\cos x =  - \dfrac{{\sqrt 6 }}{4}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  \pm \arccos \dfrac{{\sqrt 6 }}{4} + k2\pi \\
x =  \pm \arccos \left( { - \dfrac{{\sqrt 6 }}{4}} \right) + k2\pi 
\end{array} \right.
\end{array}\)