Đáp án:

\[\left[ \begin{array}{l}
x = k\pi \\
x =  \pm \dfrac{{2\pi }}{3} + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\]

Giải thích các bước giải:

 Ta có:

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