4cos^3(x)-(sinx+cosx)=0

[latex]4cos^3x - cosx = sqrt{1-cos^2x}\ 16cos^6x -8cos^4x +cos^2x = 1 - cos^2x\ 16cos^6x - 8cos^4x +2cos^2x -1 = 0\ cos^2x = t\ 16t^3 - 8t^2 +2t - 1 = 0\ 2t(8t^2 +1) - (8t^2+1)=0\ (8t^2 +1)(2t-1)=0\ t^2 = -frac{1}{8} t = frac{1}{2}\ net resheniy cos^2x = frac{1}{2}\ . cosx = ^+_-frac{1}{sqrt{2}}\ . x = frac{pi}{4}+frac{pi}{2}k[/latex]