cos2x=根号2sin(x+π/4)
问题描述：
cos2x=根号2sin(x+π/4)
答
cos^2x-sin^2x=sinx+cosx(cosx+sinx)(cosx-sinx)=cosx+sinx(cosx+sinx)(cosx-sinx-1)=0cosx+sinx=0或sinx-cosx=-1根号2sin(x+pi/4)=0或根号2sin(x-pi/4)=-1所以x=kpi-pi/4或x=2kpi或2kpi-pi/2