$\begin{array}{l}a)\,\,\cos3x - \cos2x - \cos x - 1 = 0\\ \Leftrightarrow (\cos3x - \cos x) + 1 - 2\sin^2x - 1 = 0\\ \Leftrightarrow -2\sin2x\sin x - 2\sin^2x = 0\\ \Leftrightarrow \sin x(\sin2x + \sin x) = 0\\ \Leftrightarrow \left[\begin{array}{l}\sin x = 0\\\sin2x = \sin(-x)\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = k\pi\\2x = - x + k2\pi\\2x = \pi + x + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = k\pi\\x = k\dfrac{2\pi}{3}\\x = \pi +k2\pi\end{array}\right.\quad (k \in \Bbb Z)\\ b)\,\,\cos^23x\cos2x - \cos^2x = 0\\ \Leftrightarrow (1 + \cos6x)\cos2x - (1 + \cos2x) = 0\\ \Leftrightarrow \cos6x\cos2x - 1 = 0\\ \Leftrightarrow (4\cos^32x - 3\cos2x)\cos2x - 1 = 0\\ \Leftrightarrow 4\cos^42x - 3\cos^22x - 1 = 0\\ \Leftrightarrow \left[\begin{array}{l}\cos^22x = - \dfrac{1}{4}\quad (loại)\\\cos^22x = 1 \qquad (nhận)\end{array}\right.\\ \Leftrightarrow \cos2x = \pm 1\\ \Leftrightarrow 2x = k\pi\\ \Leftrightarrow x = k\dfrac{\pi}{2}\quad (k \in \Bbb Z) \end{array}$
