y=sinx^2+cosx^2+2sinxcosx+2cosx^2
= 1+sin2x+1+cos2X
=2+根号2*sin(2x+π/4)
单调递减,则需要2x+π/4∈(2kπ+π/2,2kπ+3*π/2)
区间就出来了
(2)最大值,当sin(2x+π/4)取到1的时候,为2+根号2
最小值,当sin(2x+π/4)取到-1的时候,为2-根号2
y=sinx^2+cosx^2+2sinxcosx+2cosx^2
= 1+sin2x+1+cos2X
=2+根号2*sin(2x+π/4)
单调递减,则需要2x+π/4∈(2kπ+π/2,2kπ+3*π/2)
区间就出来了
(2)最大值,当sin(2x+π/4)取到1的时候,为2+根号2
最小值,当sin(2x+π/4)取到-1的时候,为2-根号2