(1)f(x)=sin2x-2sin²x
=sin2x+cos2x-1 =√2sin(2x+π/4)-1最小正周期T=π(2)令-π/2+2kπ≤2x+π/4≤π/2+2kπ
k∈Z得:-3π/8+kπ≤x≤π/8+kπ
k∈Z∴函数f(x)的单调递增区间为:【-3π/8+kπ,π/8+kπ】
(2)f(x)=0
sin(2x+π/6)=-1/2所以2x+π/6=2kπ-5π/6,2x+π/6=2kπ-π/6
x=kπ-π/2,x=kπ-π/6所以是{x|x=kπ-π/2,x=kπ-π/6,k∈Z}