函数f(x)=sinx减根号3cosx(x属于[负派,0])的单调递增区间是?急

问题描述:

函数f(x)=sinx减根号3cosx(x属于[负派,0])的单调递增区间是?急


f(x)
=sinx-√3cosx
=2(sinx•1/2-cosx•√3/2)
=2sin(x-π/3)
由:-π/2+2kπ≤x-π/3≤π/2+2kπ
得:-π/6+2kπ≤x≤5π/6+2kπ
∵x∈[-π,0]
取交集,得:[-π/6,0]