1,f(x)=2√3sinxcosx+2(cosx)^2-1=√3sin2x+cos2x=2sin(2x+π/6)
所以最小正周期是2π/2=π；
2,2kπ-π/2≦2x+π/6≦2kπ+π/2；解得：kπ-π/3≦x≦kπ+π/6;
所以单调增区间是：[kπ-π/3,kπ+π/6],k属于Z.
3,从单调区间取k=0可看出函数在[-π/3,π/6]上单增,在接下来[π/6,π/3]是单减,所以函数在[-π/6,π/3]上最大值在x=π/6处取得,最小值要比较x=-π/6,x=π/3两点的值.
最大值：f(π/6)=2;
f(-π/6)=-1；f(π/3)=1;
所以最小值为f(-π/6)=-1