y=sin²+cos²x+2sinxosx+1+cos2x
=sin2x+cos2x+2
=√2sin(2x+/4)+2
T=2π/2=π详细过程，谢啦o(∩_∩)osin2x+cos2x =√2(√2/2*sin2x+√2/2cos2x) =√2(sin2xcosπ/4+cos1xsinπ/4) =√2sin(2x+π/4)最值详细过程算我白做就是当x=什么时它有最大最小值OK哦，对不起，没看到 2x+π/4=2kπ+π/2,x=kπ+π/8,最大值=√2+2 2x+π/4=2kπ-π/2,x=kπ-3π/8,最小值=-√2+2