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]