f（x）=(√3)sin2x-2sin²x ；（1）.求f（x）最大值 （2）.求f（x）的零点的集合
(1)f(x)=(√3)sin2x+cos2x-1=2[(√3/2)sin2x+(1/2)cos2x]-1=2[sin2xcos(π/6)+cos2xsin(π/6)]-1
=2sin(2x+π/6)-1
-2≦2sin(2X+π/6)≦2,-3≦2sin(2x+π/6)-1≦1
即-3≦f(x)≦1,故f(x)的最大值是1,最小值是-3.
(2)令f(x)=2sin(2x+π/6)-1=0
得sin(2x+π/6)=1/2,故2x+π/6=π/6+2kπ,x=kπ,k∈Z.