解下列方程 cos2x=cosx+sinx sin^4x-cos^4x=cosx+sinx

1)cos2x=cosx+sinx --->cos^2 x-sin^2 x=cosx+sinx---> (cosx+sinx)(cosx-sinx-1)=0---> cosx+sinx=0---> tgx=-1--> x= kπ-π/4,这里k为任意整数.orcosx-sinx-1=0--> cos(x+π/4)=1/√2--> x+π/4=2kπ+/-π/...