y'=(0+cosx)(1+cosx)+(1+sinx)(0-sinx)
=cosx+cos²x-sinx-sin²x
=(cosx-sinx)+(cosx+sinx)(cosx-sinx)
=(cosx-sinx)(cosx+sinx+1)
和x轴平行则k=0
所以y'=0
所以cosx=sinx,cosx+sinx=-1
所以tanx=1,sin(x+π/4)=-√2/2
x=kπ+π/4,x+π/4=2kπ-3π/4,x+π/4=2kπ-π/4
x=kπ+π/4,x=2kπ-π,x=2kπ-π/2
x=kπ+π/4,即x=2kπ+π/4,x=2kπ+5π/4
x=2kπ+π/4.y=3/2+√2
x=2kπ+5π/4,y=3/2-√2
x=2kπ-π,y=0
x=2kπ-π/2,y=0
所以(2kπ+π/4,3/2+√2),(2kπ+5π/4,3/2-√2),(2kπ-π,0),(2kπ-π/2,0)