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.
or cosx-sinx-1=0--> cos(x+π/4)=1/√2--> x+π/4=2kπ+/-π/4--> x=2kπ or 2kπ-π/2
2)sin^4x-cos^4x=cosx+sinx--->(sinx^2+cosx^2)(sinx+cosx)(sinx-cosx)=sinx+cosx-->(sinx+cosx)(sinx-cosx-1)=0
-->
sinx+cosx=0--->x= kπ-π/4
or sinx-cosx-1=0-->cos(x+π/4)=-1/√2--> x+π/4=(2k+1)π+/-π/4--> x=(2k+1)π or (2k+1)π-π/2