Đáp án:
\[x = \pm \dfrac{\pi }{6} + \dfrac{{k\pi }}{2}\,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\cos 4x = 2{\cos ^2}2x - 1 \Rightarrow {\cos ^2}2x = \dfrac{{\cos 4x + 1}}{2}\\
{\cos ^2}2x = \dfrac{1}{4}\\
\Leftrightarrow \dfrac{{1 + \cos 4x}}{2} = \dfrac{1}{4}\\
\Leftrightarrow 1 + \cos 4x = \dfrac{1}{2}\\
\Leftrightarrow \cos 4x = - \dfrac{1}{2}\\
\Leftrightarrow 4x = \pm \dfrac{{2\pi }}{3} + k2\pi \\
\Leftrightarrow x = \pm \dfrac{\pi }{6} + \dfrac{{k\pi }}{2}\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
Vậy \(x = \pm \dfrac{\pi }{6} + \dfrac{{k\pi }}{2}\,\,\,\,\left( {k \in Z} \right)\)
