求函数y=sinx^4+2√3sinxcosx-cosx^4的最小正周期

y=sinx^4+2√3sinxcosx-cosx^4=[(sinx)^2+(cosx)^2][(sinx)^2-(cosx)^2]+2√3sinxcosx=1*[(sinx)^2-(cosx)^2]+2√3sinxcosx=-cos2x+√3sin2x=sin(2x-π/6)最小正周期T=2π/2=π最小值为-1,当2x-π/6=(-3/2)π+2kπ,k∈z即x=-2π/3+kπ时取最小单调增:-π/2+2kπ<2x-π/6<π/2+2kπ -π/6+kπ<x<π/3+kπ在[0,π]范围内,k取00<x<π/3即当x∈(0,π/3)时函数是单调递增的 收起