1、
f(x)=-(cos²x-sin²x)+√3(2sinxcosx)
=√3sin2x-cos2x
=2sin(2x-π/6)
T=2π/2=π
递减则2kπ+π/2<2x-π/6<2kπ+3π/2
2kπ+2π/3<2x<2kπ+5π/3
kπ+π/3减区间(kπ+π/3,kπ+5π/6)
2、
π/12<=x<=25π/36
0<=2x-π/6<=11π/9
所以x=25π/36,最小值=2sin(11π/9)=-2sin2π/9
x=π/3,最大值=2