y=1+2sinxcosx+2(cosx)^2=1+sin2x+1+cos2x=√2*sin(2x+π/4)+2 ,因此,当 2x+π/4=3π/2+2kπ ,即 x=5π/8+kπ 时,函数取最小值 2-√2 ；当 2x+π/4=π/2+2kπ ,即 x=π/8+kπ 时,函数取最大值 2+√2 .（以上 k∈Z）...