已知函数y=(sinx+cosx)^2+2cos^x

y=(sinx+cosx)^2+2cos^x
=2+sin2x+cos2x
=2+√2sin(2x+π/4)
y max=2+√2,
y min=2-√2.
2kπ+π/2≤2x+π/4≤2kπ+3π/2
2kπ+π/4≤2x≤2kπ+5π/4
kπ+π/8≤x≤kπ+5π/8
它的递减区间[kπ+π/8,kπ+5π/8],k∈Z.

y=(sinx+cosx)^2+2(cosx)^2
=(sinx)^2+(cosx)^2+2sinxcosx+2(1+cos2x)/2
=1+sin2x+1+cos2x
=2+sin(2x+π/4)
-1<=sin(2x+π/4)<=1
1<=y<=3
sinx在[π/2+2kπ,3π/2+2kπ]上递减
所以y递减区间
π/2+2kπ<=2x+π/4<=3π/2+2kπ
π/8+kπ<=x<=5π/8+kπ
k是整数