问题描述: 已知f(x)=√3sin(x+θ)+cos(x-θ)是偶函数,0<θ<x,求θ是0<θ<π 1个回答 分类:数学 2014-12-11 问题解答: 我来补答 f(x)=√3sin(x+θ)+cos(x-θ)是偶函数,即f(-x)=f(x)f(-x)=√3sin(-x+θ)+cos(-x-θ)=-√3sin(x-θ)+cos(x+θ)=√3sin(x+θ)+cos(x-θ)即:cos(x+θ)-cos(x-θ)=√3sin(x+θ)+√3sin(x-θ)即:-2sinxsinθ=2√3sinxcosθ,故tanθ=-√3所以θ=2π/3 展开全文阅读