Алгебра: cosx-sinx=4cosx*sin^2x

Решите задачу:

$cosx-sinx=4cosx*sin^2x\\ cosx(1-4sin^2x)=sinx\\ cosx(1-4(1-cos^2x))=sinx\\ cosx(4cos^2x-3)=sinx\\ cos^2x(4cos^2x-3)^2=sin^2x\\ cos^2x(4cos^2x-3)^2=1-cos^2x\\ cos^2x=t\\ t(4t-3)^2=1-t\\ (2t-1)(8t^2-8t+1)=0\\ t=\frac{1}{2}\\ t=-\frac{\sqrt{2}-2}{4}\\ t=\frac{\sqrt{2}+2}{4}\\\\ cosx=+-\frac{\sqrt{2}}{2}\\ x=-\frac{\pi}{4}+\pi*n\\ x=\frac{-3\pi}{8}+\pi*n\\ x=\frac{-7\pi}{8}+\pi*n$