Giải phương trình \[\cos 2x + \cos 4x + \cos 6x = \cos x\cos 2x\cos 3x + 2\]

\[cos2x + cos4x + cos6x = cosxcos2xcos3x + 2\]

\[ \Leftrightarrow 2cos4xcos2x + cos4x = \frac{1}{2}cos2x(cos4x + cos2x) + 2\]

\[ \Leftrightarrow 2cos4xcos2x + cos4x = \frac{1}{2}cos2xcos4x + \frac{1}{2}co{s^2}2x + 2\]

\[ \Leftrightarrow \frac{3}{2}cos4xcos2x + cos4x = \frac{1}{2}co{s^2}2x + 2\]

\[ \Leftrightarrow 3cos4xcos2x + 2cos4x = co{s^2}2x + 4\]

\[ \Leftrightarrow 3(2co{s^2}2x - 1)cos2x + 2(2co{s^2}2x - 1) = co{s^2}2x + 4\]

\[ \Leftrightarrow 6co{s^3}2x - 3cos2x + 4co{s^2}2x - 2 = co{s^2}2x + 4\]

\[ \Leftrightarrow 6co{s^3}2x + 3co{s^2}2x - 3cos2x - 6 = 0\]

\[ \Leftrightarrow 2co{s^3}2x + co{s^2}2x - cos2x - 2 = 0\]

\[ \Leftrightarrow 2(co{s^3}2x - 1) + cos2x(cos2x - 1) = 0\]

\[ \Leftrightarrow 2(cos2x - 1)(co{s^2}2x + cos2x + 1) + cos2x(cos2x - 1) = 0\]

\[ \Leftrightarrow (cos2x - 1)(2co{s^2}2x + 2cos2x + 2 + cos2x) = 0\]

\[ \Leftrightarrow (cos2x - 1)(2co{s^2}2x + 3cos2x + 2) = 0\]

\[ \Leftrightarrow cos2x = 1 \Leftrightarrow 2x = k2\pi \Leftrightarrow x = k\pi (k \in \mathbb{Z})\]

Vậy nghiệm của phương trình đã cho là: \[x = k\pi \,\,\left( {k \in \mathbb{Z}} \right)\]