Ta có
$\cos x + \cos(2x) + \cos(3x) + \cos(4x) = 0$
$<-> \cos x + \cos(3x) + \cos(2x) + \cos(4x) = 0$
$<-> 2\cos(2x) \cos x + 2 \cos(3x) \cos x = 0$
$<-> \cos x[\cos(2x) + \cos(3x)] = 0$
Vậy $\cos x = 0$ hoặc $\cos (2x) + \cos(3x) = 0$
TH1: $\cos x = 0$
Suy ra $x = \dfrac{\pi}{2} + k\pi$
TH2: $\cos(2x) = -\cos(3x)$
Suy ra
$\cos(2x) = \cos (\pi - 3x)$
$<-> 2x = \pi - 3x + 2k\pi$ hoặc $2x = 3x - \pi + 2k\pi$
$<-> x = \dfrac{\pi}{5} + \dfrac{2k\pi}{5}$ hoặc $x = \pi + 2k\pi$.
