已知向量a=(2sinx,√3cosx),b=(sinx,2sinx),函数f(x)=a·b

(1)
f(x)=a·b=2sin²x+2√3sinxcosx
=1-cos2x+√3sin2x
=2(√3/2*sin2x-1/2*cos2x)+1
=2sin(2x-π/6)+1
由2kπ-π/2≤2x-π/6≤2kπ+π/2
得 kπ-π/6≤x≤kπ+π/3
∴f(x)的单调递增区间是
[kπ-π/6,kπ+π/3 ],k∈Z
(2)
不等式f(x)≥m对x∈[0,兀/2]都成立
则f(x)min≥m
∵x∈[0,兀/2]
∴2x-π/6∈[-π/6,5π/6]
∴2x-π/6=-π/6,即x=0时,
f(x)min=2sin(-π/6)+1=0
∴m≤0
∴实数m的最大值是0