题目

求函数f(x)=2sin(π-x)sin(π/2-x)+2根号3sin^2x-根号3的单调递减区间

提问时间：2020-10-16

答案

f(x)=2sinxcosx+2√3sin²x-√3
=2sinxcosx+√3sin²x+√3(sin²x-1)
=2sinxcosx+√3sin²x-√3cos²x
=2sinxcosx+√3(sin²x-cos²x)
=sin2x-√3cos2x
=2(1/2 sin2x-√3/2 cos2x)
=2sin(2x-π/3)
sinx的单调递减区间为[π/2+2nπ,3π/2+2nπ] ,n为整数
π/2+2nπ≤2x-π/3≤3π/2+2nπ
5π/6+2nπ≤2x≤11π/6+2nπ
5π/12+nπ≤x≤11π/12+nπ
所以sin(2x-π/3)的单调递减区间为[5π/12+nπ,11π/12+nπ],即
函数f(x)=2sin(2x-π/3)的单调递减区间为[5π/12+nπ,11π/12+nπ],n∈Z