2cos^2x+cos3x+cos4x=1

Решите задачу:

$2cos^{2} x + cos 3x + cos 4x = 1 \\ \frac{2(1 +cos 2x)}{2} + cos 3x + cos 4x = 1 \\ 1 + cos 2x + cos 4x + cos 3x -1 = 0 \\ cos 2x + cos 4x + cos 3 x = 0 \\ 2 cos \frac{2x - 4x}{2} cos \frac{2x + 4x}{2} + cos 3x = 0 \\ 2 cos x cos 3x + cos 3x = 0 \\$
$cos 3x (2cos x +1) = 0 \\ 1) cos 3x = 0 \\ cos t = 0 \\ t = \frac{ \pi }{2} + \pi k \\ 3x = \frac{ \pi }{2} + \pi k \\ x_{1} = \frac{ \pi }{6} + \frac{ \pi k}{3} \\ 2) 2 cosx + 1 = 0 \\ cosx = - \frac{1}{2} \\ x_{2} = +- \frac{2 \pi }{3} + 2\pi k$