Đáp án:
$\begin{array}{l}
a)\cos 3x + \cos 2x - \cos x - 1 = 0\\
\Rightarrow 4.{\cos ^3}x - 3\cos x + 2{\cos ^2}x - 1 - \cos x - 1 = 0\\
\Rightarrow 4{\cos ^3}x + 2{\cos ^2}x - 4\cos x - 2 = 0\\
\Rightarrow \left( {2\cos x + 1} \right)\left( {2{{\cos }^2}x - 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\cos x = - \dfrac{1}{2}\\
{\cos ^2}x = 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\cos x = - \dfrac{1}{2}\\
\cos x = 1\\
\cos x = - 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{{2\pi }}{3} + k2\pi \\
x = - \dfrac{{2\pi }}{3} + k2\pi \\
x = k\pi
\end{array} \right.\\
b)\sin 3\left( {x + \pi } \right) - \sin 2\left( {x + 2\pi } \right) - \sin \left( {x + 3\pi } \right) = 0\\
\Rightarrow \sin \left( {3x + 3\pi } \right) - \sin \left( {2x + 4\pi } \right) - \sin \left( {x + 3\pi } \right) = 0\\
\Rightarrow - \sin 3x - \sin 2x + \sin x = 0\\
\Rightarrow - 3\sin x + 4{\sin ^3}x - 2.\sin x.\cos x + \sin x = 0\\
\Rightarrow \sin x.\left( {4{{\sin }^2}x - 2\cos x - 2} \right) = 0\\
\Rightarrow 2.\sin x.\left( {2{{\sin }^2}x - \cos x - 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\sin x = 0\\
2.\left( {1 - {{\cos }^2}x} \right) - \cos x - 1 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\sin x = 0\\
2{\cos ^2}x + \cos x - 1 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\sin x = 0\\
\left( {2\cos x - 1} \right)\left( {\cos x + 1} \right) = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\sin x = 0\\
\cos x = - 1\\
\cos x = \dfrac{1}{2}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{\pi }{3} + k2\pi \\
x = - \dfrac{\pi }{3} + k2\pi
\end{array} \right.
\end{array}$
