解f（x）=√3 sin2x +2cos^2x+a
=√3 sin2x +2cos^2x-1+a+1
=√3 sin2x +cos2x+a+1
=2（√3 /2sin2x +1/2cos2x）+a+1
=2sin(2x+π/6)+a+1
由x屬于[0,π/2]
則2x屬于[0,π]
則2x+π/6屬于[π/6,7π/6]
故當(dāng)2x+π/6=7π/6時(shí),函數(shù)f（x）=2sin(2x+π/6)+a+1在[0,π/2]上
有最小值2×（-1/2)+a+1=a
而由題知f（x）=√3 sin2x +2cos²x+a,在[0,π/2]上最小值-1
則a=-1.