可化简：f(x)=2cos²x+根号3sin2x+a
=1+cos2x+根号3sin2x+a
=2(sin2xcosπ/6+cos2xsinπ/6)+a+1
=2sin(2x+π/6)+a+1
（1）若x∈R时,最小正周期T=2π/2=π.
（2）x∈[0,π/2]时,则(2x+π/6）∈[π/6,7π/6],
此时sin(2x+π/6)∈[-1/2,1],
所以2*1+a+1=4 则a=1