Cos 5x+ cos 2x+ cos3x + cos4x = 0

$(cos5x+cos3x)+(cos2x+cos4x)=0\\ 2cos4x*cosx+2cos3x*cosx=0\\ cosx(2cos4x+2cos3x)=0\\ \left \{ {{cosx=0} \atop {cos4x+cos3x=0}} \right. \\ \\ \left \{ {{x_{1}=\pi \*n-\frac{\pi}{2}} \atop {cos^4x+sin^4x-6sin^2xcos^2x+cos^3x-3sin^2xcosx=0}} \right. \\ \\ cos^4x+(1-cos^2x)^2-6(1-cos^2x)cos^2x+cos^3x-3(1-cos^2x)cosx=0\\ cosx=t\\ 8t^4+4t^3-8t^2-3t+1=0\\ (t+1)(8t^3-4t^2-4t+1)=0\\ t=-1\\ cosx=-1\\ x_{2}=2\pi\ n+\pi\\ \\$
$cos4x+cos3x=0\\ 2cos\frac{7x}{2}*cos\frac{x}{2}=0\\ cos\frac{7x}{2}=0\\ x=-\frac{\pi}{7}-\frac{2\pi*n}{7}\\ cos\frac{x}{2}=0\\ x=2\pi+\pi$