解;f(x)=cos^2x-√3sinxcosx-1/2=1/2*(1+cos2x)- √3/2sin2x-1/2
= -(sin(2x-π/6))
x∈[0,π/2], 2x-π/6∈[-π/6,5π/6]
-sin(2x-π/6)的最小值=-1,此时2x-π/6=π/2,x=π/3