Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
\cos 3x + \cos 4x + \cos 5x + \cos 2x = 0\\
 \Leftrightarrow \left( {\cos 5x + \cos 3x} \right) + \left( {\cos 4x + \cos 2x} \right) = 0\\
 \Leftrightarrow 2.\cos \dfrac{{5x + 3x}}{2}.\cos \dfrac{{5x - 3x}}{2} + 2.cos\dfrac{{4x + 2x}}{2}.\cos \dfrac{{4x - 2x}}{2} = 0\\
 \Leftrightarrow 2.\cos 4x.\cos x + 2\cos 3x.\cos x = 0\\
 \Leftrightarrow 2\cos x.\left( {\cos 4x + \cos 3x} \right) = 0\\
 \Leftrightarrow 2\cos x.2.\cos \dfrac{{4x + 3x}}{2}.\cos \dfrac{{4x - 3x}}{2} = 0\\
 \Leftrightarrow \cos x.\cos \dfrac{{7x}}{2}.\cos \dfrac{x}{2} = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\cos \dfrac{{7x}}{2} = 0\\
\cos \dfrac{x}{2} = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
\dfrac{{7\pi }}{2} = \dfrac{\pi }{2} + k\pi \\
\dfrac{x}{2} = \dfrac{\pi }{2} + k\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = \dfrac{\pi }{7} + \dfrac{{2k\pi }}{7}\\
x = \pi  + k2\pi 
\end{array} \right.\\
2,\\
\cos 2x + 3\cos x - 4 = 0\\
 \Leftrightarrow \left( {2{{\cos }^2}x - 1} \right) + 3\cos x - 4 = 0\\
 \Leftrightarrow 2{\cos ^2}x + 3\cos x - 5 = 0\\
 \Leftrightarrow \left( {\cos x - 1} \right)\left( {2\cos x + 5} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x = 1\\
\cos x =  - \dfrac{5}{2}\,\,\,\,\,\left( {L,\,\,\, - 1 \le \cos x \le 1} \right)
\end{array} \right.\\
 \Leftrightarrow \cos x = 1\\
 \Leftrightarrow x = k2\pi \\
3,\\
{\cos ^2}x + {\cos ^2}2x + {\cos ^2}3x + {\cos ^2}4x = 0\\
{\cos ^2}x \ge 0;\,\,\,{\cos ^2}2x \ge 0,\,\,\,\,{\cos ^2}3x \ge 0,\,\,\,{\cos ^2}4x \ge 0\\
 \Rightarrow \left\{ \begin{array}{l}
\cos x = 0\\
\cos 2x = 0\\
\cos 3x = 0\\
\cos 4x = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
2x = \dfrac{\pi }{2} + k\pi \\
3x = \dfrac{\pi }{2} + k\pi \\
4x = \dfrac{\pi }{2} + k\pi 
\end{array} \right.\,\,\,\,\,\left( {vn} \right)
\end{array}\)