(1)∵f(x)=4cosxsin(x+ π 3 )- 3 =2cosx( 1 2 sinx+ 3 2 cosx)- 3 =sin2x+ 3 (1+cos2x)- 3 =sin2x+ 3 cos2x=2sin(2x+ π 3 )∴函数的最小正周期T= 2π 2 =π.由2kπ+ π 2 ≤2x+ π 3 ≤2kπ+ 3π 2 ,k∈Z,解得kπ+ π 12 ≤x≤kπ+