f(x)=√3sinxcosx+sin²x-1/2
=√3/2*(2sinxcosx)-1/2*(1-2sin²x)
=√3/2*sin2x-1/2*cos2x
=sin(2x-π/6)
(1) 令2x-π/6=kπ+π/2,得：x=kπ/2+π/3
所以对称轴方程为：x=kπ/2+π/3 (k∈Z)
(2) ∵0≤x≤π/2,∴-π/6≤2x-π/6≤5π/6
∴-1/2≤sin(2x-π/6)≤1
∴f(x)max=1,f(x)min=-1/2
已知函数f（x）=√3sinxcosx+sin^2x-1/2 求
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