244.решите неравенство f'(x)<0

Решите задачу:

$f'(x)=\cos x +\cos 2x+\cos 3x\ \textless \ 0;$
$\cos x+2\cos^2 x-1+4\cos^3x-3\cos x\ \textless \ 0;$
$4\cos^3x+2\cos^2 x-2\cos x-1\ \textless \ 0;$
$(2\cos^2x-1)(2\cos x+1)\ \textless \ 0;$
$(\cos x-\frac{\sqrt{2}}{2})(\cos x+\frac{\sqrt{2}}{2})(\cos x+\frac{1}{2})\ \textless \ 0;$
$\cos x\in(-1; -\frac{\sqrt{2}}{2})\cup(-\frac{1}{2};\frac{\sqrt{2}}{2})$
$x\in(\frac{\pi}{4}+2\pi n; \frac{2\pi}{3}+2\pi n)\cup(\frac{3\pi}{4}+2\pi n; \frac{5\pi}{4}+2\pi n)\cup$
$\cup(\frac{4\pi}{3}+2\pi n; \frac{7\pi}{4}+2\pi n),\,n\in\mathbb{Z}$