1f(x)=sinx+√3cosx=2(1/2sinx+√3/2cosx)=2(cosπ/3sinx+sinπ/3cosx)=2sin(x+π/3)x+π/3∈[-π/2+2kπ,π/2+2kπ] 即 x∈[-5π/6+2kπ,π/6+2kπ] 單調(diào)增x+π/3∈[π/2+2kπ,3π/2+2kπ] 即 x∈[π/6+2kπ,7π/6+...請問=2(1/2sinx+√3/2cosx) =2(cosπ/3sinx+sinπ/3cosx)怎么得的π/3=60º cos60º=1/2sin60º=√3/2不懂可追問 先采納哦親