问题描述: 已知函数f(x)=cos(2x-π/3)+sin^2x-cos^2x,设函数g(x)=[f(x)]^2+f(x),求g(x)值域 1个回答 分类:综合 2014-11-16 问题解答: 我来补答 ∵f(x)=cos(2x-π/3)+(sinx)^2-(cosx)^2=cos(2x-π/3)-cos2x=2sin(2x-π/6)sin(π/6)=sin(2x-π/6).∴g(x)=[sin(2x-π/6)]^2+sin(2x-π/6)=[sin(2x-π/6)+1/2]^2-1/4.∴当sin(2x-π/6)=-1/2时,g(x)有最小值为-1/4; 当sin(2x-π/6)=1时,g(x)有最大值为(1+1/2)^2-1/4=9/4-1/4=2.∴g(x)的值域是[-1/4,2]. 展开全文阅读