函数y=2sinx(sinx+cosx)
＝2sinxsinx+2sinxcosx
=sin2x－cos2x+1
=√2sin(2x-π/4)+1
1.求此函数的最小正周期
T=2π/2=π
2.当X取什么值时,此函数有最大值
2x-π/4=2kπ+π/2
x=kπ+3π/8,k是整数,y max=√2+1
3.求此函数的单调递减区间
π/2+2kπ