Đáp án:

\[x = \dfrac{\pi }{2} + k\pi \,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {k \in Z} \right)\]

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
\cos x.\cos y = \dfrac{1}{2}.\left[ {\cos \left( {x + y} \right) + \cos \left( {x - y} \right)} \right]\\
\cos 2x = 2{\cos ^2}x - 1 \Rightarrow {\cos ^2}x = \dfrac{{\cos 2x + 1}}{2}\\
\cos 5x.\cos x = \cos 4x.\cos 2x + 3{\cos ^2}x + 1\\
 \Leftrightarrow \dfrac{1}{2}.\left[ {\cos \left( {5x + x} \right) + \cos \left( {5x - x} \right)} \right] = \dfrac{1}{2}.\left[ {\cos \left( {4x + 2x} \right) + \cos \left( {4x - 2x} \right)} \right] + 3{\cos ^2}x + 1\\
 \Leftrightarrow \dfrac{1}{2}\cos 6x + \dfrac{1}{2}\cos 4x = \dfrac{1}{2}\cos 6x + \dfrac{1}{2}\cos 2x + 3{\cos ^2}x + 1\\
 \Leftrightarrow \dfrac{1}{2}\cos 4x - \dfrac{1}{2}\cos 2x - 3{\cos ^2}x - 1 = 0\\
 \Leftrightarrow \dfrac{1}{2}\cos 4x - \dfrac{1}{2}\cos 2x - 3.\dfrac{{\cos 2x + 1}}{2} - 1 = 0\\
 \Leftrightarrow \cos 4x - \cos 2x - 3.\left( {\cos 2x + 1} \right) - 2 = 0\\
 \Leftrightarrow \left( {2{{\cos }^2}2x - 1} \right) - \cos 2x - 3\cos 2x - 5 = 0\\
 \Leftrightarrow 2{\cos ^2}2x - 4\cos 2x - 6 = 0\\
 \Leftrightarrow {\cos ^2}2x - 2\cos 2x - 3 = 0\\
 \Leftrightarrow \left( {\cos 2x - 3} \right)\left( {\cos 2x + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos 2x = 3\\
\cos 2x =  - 1
\end{array} \right.\\
 \Leftrightarrow \cos 2x =  - 1\,\,\,\,\,\,\,\,\,\,\,\,\left( { - 1 \le \cos 2x \le 1} \right)\\
 \Leftrightarrow 2x = \pi  + k2\pi \\
 \Leftrightarrow x = \dfrac{\pi }{2} + k\pi \,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)