Question

已知函数f(x)=2cosxsin(x+π/3)-根号3sin^2x+sinxcosx+2(x∈R),

1,求函数f(x)的最小正周期和单调增区间；

Answer 1

f(x)=sinxcosx+sqrt(3)(cosx)^2-sqrt(3)(sinx)^2+sinxcosx+2
=2sinxcosx+sqrt(3)[(cosx)^2-(sinx)^2]
=sin2x+sqrt(3)cos2x
=2sin(2x+π/3)
所以最小正周期是π
单调增区间[-5π/12+kπ，π/12+kπ]，k是整数。

Answer 2

原式=2cosx(sinx/2+√3*sinx/2)-√3*(sinx)^2+sinxcosx+2
=2sinxcosx+√3（cos^2(x)-sin^2(x)）+2
=sin2x+√3cos2x+2
=2sin(2x+π/3)+2
所以最小正周期是π，单调增区间为[-5π/12+kπ，π/12+kπ]，k为整数