f(x)=3sinxcosx+√3cos²x-√3/2
= 3/2 sin2x+√3/2cos2x
= √3(√3/2 sin2x+1/2cos2x)
=√3 sin(2x+π/6)
(1) 最小正周期 T=2π/2=π;
(2)递减区间满足2kπ+π/2≦2x+π/6≦2kπ+3π/2
即减区间为[kπ+π/6,kπ+2π/3]
(3) x∈[0,π/2], 2x+π/6∈[π/6,7π/6], sin(2x+π/6)∈[-1/2,1],
最小值为 -√3/2,这时 x=π/2;
最大值为 √3,这时 2x+π/6=π/2 即 x=π/3