f(x)=2sin(2x+π/6)-4cos^2 x+2=2sin(2x-π/6)
T=2π/2=π
∴函数f(x)的周期为π还有一问答了加分~~：若x∈[π/4，π/2]求fx 的值域和单调区间~∵f(x)=2sin(2x-π/6)∴单调增区间：2kπ-π/2<=2x-π/6<=2kπ+π/2==>kπ-π/6<=x<=kπ+π/3∵x∈[π/4，π/2]∴x∈[π/4，π/3]时，单调增，x∈[π/3，π/2]时，单调减f(π/4)=2sin(π/2-π/6)=√3，f(π/2)=2sin(π-π/6)=1值域为[1，2]