向量a=(sinx,sinx),b=(√3cosx,sinx)
函数f(x)=2ab-m =2√3sinxcosx+2sin²x-m
=√3 sin2x+1-cos2x-m
=2(√3/2sin2x-1/2cos2x)+1-m
=2sin(2x-π/6)+1-m
f(x)的最小正周期T=2π/2=π
由 2kπ-π/2≤2x-π/6≤2kπ+π/2,k∈Z
得：kπ-π/6≤x≤kπ+π/3,k∈Z
∴f(x)单调递增区间是[kπ-π/6,kπ+π/3],k∈Z
(2)
f(x)=0得 2sin(2x-π/6)+1-m=0
sin(2x-π/6)=(m-1)/2
x∈ [0,π/2]时,2x-π/6∈[-π/6,5π/6]
若f(x)=0有两个两个不同的实根x1 x2
需2x1-π/6∈[π/6,π/2)
2x2-π/6∈(π/2,5π/6]
∴sin(2x-π/6)∈[1/2,1)
∴ 1/2≤(m-1)/2