1 g(x)=b²-1=(1,sin2x)(1,sin2x)-1=1+sin²2x-1=sin²2x
∴g(x)=0 => sin²2x=0 => 2x=kπ => x=kπ/2,k为任意整数
∴g(x)=0的解集为{kπ/2:k∈Z}
2 f(x)=ab-1=(2cos²x,√3)(1,sin2x)-1=2cos²x+√3sin2x-1
=cos2x+√3sin2x=2sin(2x+π/6)
∴f(x)的最小正周期为2π/2=π,单调递增区间为
2kπ-π/2≤2x+π/6≤2kπ+π/2 => 2kπ-2π/3≤2x≤2kπ+π/3
=> kπ-π/3≤x≤kπ+π/6,即[kπ-π/3,kπ+π/6]