Đáp án:
\(x = \frac{\pi }{2} + k\pi \,\,\,\left( {k \in Z} \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
\cos 4x - \cos 2x = 2\\
\Leftrightarrow 2{\cos ^2}2x - 1 - \cos 2x - 2 = 0\\
\Leftrightarrow 2{\cos ^2}2x - \cos 2x - 3 = 0\\
\Leftrightarrow \left( {2\cos 2x - 3} \right)\left( {\cos 2x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2\cos 2x = 3\\
\cos 2x = - 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos 2x = \frac{3}{2}\,\,\left( {ktm} \right)\\
\cos 2x = - 1
\end{array} \right.\\
\Leftrightarrow 2x = \pi + k2\pi \Leftrightarrow x = \frac{\pi }{2} + k\pi \,\,\,\left( {k \in Z} \right).
\end{array}\)
