Question

cosx-sinx=4cosx*sin^2x

Answer 1

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