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

Решите задачу:

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