已知函数f(x)=根号3sin(2x+fai)-cos(2x+fai)(0

（1）f(x)=√3sin(2x+φ)-cos(2x+φ）=2[√3/2*sin(2x+φ)-1/2*cos(2x+φ）]
=2sin(2x+φ-π/6) 因为是偶函数 ∴函数f(x)在x=0处取最大值或最小值
∴2*0+φ-π/6=π/2 +kπ φ= 2π/3 +kπ 又∵0