解f（x）=2cos²x+根號3sin2x+a
=2cos²x-1+√3sin2x+a+1
=cos2x+√3sin2x+a+1
=2（√3/2sin2x+1/2cos2x）+a+1
=2sin（2x+π/6)+a+1
由x∈【0,π/2】
知2x∈【0,π】
即2x+π/6∈【π/6,7π/6】
即-1/2≤sin（2x+π/6）≤1
即-1≤2sin（2x+π/6）≤2
即a≤2sin（2x+π/6）+a+1≤a+3
即a≤f（x）≤a+3
故f（x）的最大值為a+3,最小值為a
2 f（x）=sin(2x+π/6)+cos(2x+π/3)
=sin(2x+π/6)+sin[π/2-(2x+π/3)]
=sin(2x+π/6)+sin(π/6-2x）
=sin(2x+π/6)-sin(2x-π/6）
=sin2xcosπ/6+cos2xsinπ/6-[sin2xcosπ/6-cos2xsinπ/6]
=2cos2xsinπ/6
=cos2x
1故函數(shù)的周期T=2π/2=π
2當(dāng)2x=2kπ+π,k屬于Z時(shí),y有最小值-1
即x=kπ+π/2,k屬于Z時(shí),y有最小值-1
故函數(shù)y=f（x）的最小值為-1,對應(yīng)x的集合是｛x/x=kπ+π/2,k屬于Z｝
3由y≤0
即cos2x≤0
即2kπ+π/2≤2x≤2kπ+3π/2,k屬于Z,
即kπ+π/4≤x≤kπ+3π/4,k屬于Z
故使y≤0的x取值范圍｛x/kπ+π/4≤x≤kπ+3π/4,k屬于Z｝