Question

已知函数f(x)=2cosxsin(x+π/3)-√3sinx的平方+sinxcosx的最小正周期,单调递增区间

Answer 1

f(x)=sin(x+x+π/3)-sin(-π/3)-√3/2(1+cos2x)+1/2sin2x
=sin(2x+π/3)-√3/2cos2x+1/2sin2x
=√3/2cos2x+1/2sin2x-√3/2cos2x+1/2sin2x
=sin2x
所以T=π;增区间为(kπ-π/4,kπ+π/4)

Answer 2

2cosxsin(x+π/3)=cosxsinx+√3cos²x
f(x)=2sinxcosx+√3cos²x-√3sin²x=sin2x+√3cos2x
f(x)=2sin(2x+π/3)
T=π
增区间由-π/2+2kπ<2x+π/3<π/2+2kπ求得。

字数限制，希望能帮到你。。。

Answer 3

2cosxsin(x+π/3)=cosxsinx+√3cos²x
f(x)=2sinxcosx+√3cos²x-√3sin²x=sin2x+√3cos2x
f(x)=2sin(2x+π/3)
所以周期T=π
增区间由-5π/12+kπ<x<π/12+kπ求得。