Đáp án:
$\left[\begin{array}{l}x = \dfrac{\pi}{2} + k\pi\\x = \dfrac{\pi}{5} + k\dfrac{2\pi}{5}\\x=\pi + k2\pi\end{array}\right.\qquad (k \in \Bbb Z)$
Giải thích các bước giải:
$\begin{array}{l}\cos x + \cos2x + \cos3x + \cos4x = 0\\ \Leftrightarrow (\cos x + \cos3x)+ (\cos2x + \cos4x) = 0\\ \Leftrightarrow 2\cos2x\cos x + 2\cos3x\cos x =0\\ \Leftrightarrow \cos x(\cos2x + \cos 3x) = 0\\ \Leftrightarrow \left[\begin{array}{l}\cos x =0\\\cos2x = - \cos3x\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{2} + k\pi\\2x = \pi - 3x + k2\pi\\2x = 3x - \pi + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{2} + k\pi\\x = \dfrac{\pi}{5} + k\dfrac{2\pi}{5}\\x=\pi + k2\pi\end{array}\right.\qquad (k \in \Bbb Z) \end{array}$
