已知函数f(x)=sin(x)sin(x+π⼀2)-√3cos눀（3π+x)+1⼀2√3(x∈R)求f(x)的最小正周期？

∵f(x)=sinxsin(x+π/2)-√3cos²(3π+x)+√3/2
=sinxcosx-√3cos²x+√3/2
=(1/2)sin2x-(√3/2)cos2x
=sin(2x-π/3)
(1)∴t=2π/2=π
(2)令2x-π/3=2kπ-π/2，解得x=kπ-π/12，k∈z
令2x-π/3=2kπ+π/2，解得x=kπ+5π/12，k∈z
∴f(x)的单调递增区间为x∈[kπ-π/12，kπ+5π/12]，k∈z
(3)令2x-π/3=kπ+π/2，解得x=kπ/2+5π/12
∴f(x)的对称轴方程为x=kπ/2+5π/12，k∈z
令2x-π/3=kπ，解得x=kπ/2+π/6
∴f(x)的对称中心坐标为(kπ/2+π/6，0)，k∈z

f(x)=sin(x)sin(x+π/2)-√3cos²（3π+x)+1/2√3
=sinxcosx-√3cos²x+√3/2
=(1/2)sin2x-(√3/2)(2cos²x-1)
=(1/2)sin2x-(√3/2)cos2x
=sin(2x-π/3)
所以最小正周期T=2π/2=π
希望能帮到你O(∩_∩)O

f(x)=sin(x)sin(x+π/2)-√3cos²（3π+x)+1/2√3
=sin(x)cos(x)-√3cos²（x)+1/2√3
=(1/2)sin(2x)-(1/2)√3(1+cos2x)+1/2√3
=(1/2)sin(2x)-[(√3)/2]cos(2x)
=sin(2x)cos(π/3)-cos(2x)sin(π/3)
=sin(2x-π/3)
最小正周期为2π/2=π

f(x)=sin(x)sin(x+π/2)-√3cos²（3π+x)+1/2√3=1/2(sin2x)-√3cos²x+1/2√3
=1/2(sin2x)-√3/2* (2 cos²x-1)-√3/2+1/2√3
最小正周期T=π

最小正周期T=π。