已知f(x)=2cos^2x+根号3sin2x+a （2）若x属于【0,π/2】时,f（x）的最大值为4,求a的值（3）在(2)的条件下求满足f(x）=1且x∈【-π,π】的x的集合

f(x)=2cos²x+√3sin2x+a
=1+cos2x+√3sin2x+a
=2(√3/2*sin2x+1/2*cos2x)+a+1
=2sin(2x+π/6)+a+1
(2) 当0≤x≤π/2时,π/6≤2x+π/6≤7π/6
那么-1/2≤sin(2x+π/6)≤1
于是f(x)max=2×1+a+1=a+3=4,所以a=1
(3)f(x)=2sin(2x+π/6)+2=1
那么sin(2x+π/6)=-1/2
而-π≤x≤π,即-11π/6≤2x+π/6≤13π/6
所以2x+π/6=-π/6,或-5π/6,或7π/6,或11π/6
那么x=-π/6,或-π/2,或π/2,或5π/6
即x的集合为{-π/2,-π/6,π/2,5π/6}