题目

f(x)=根号3sin(2x-π/6)+2sin^2(x-π/12),若x∈[-π/2,π/2],求f(x)的值域
帮个忙,

提问时间：2021-04-01

答案

f(x)=√3sin(2x-π/6)+2sin²(x-π/12)
=√3sin(2x-π/6)-cos(2x-π/6)-1
=2sin(2x-π/6)cos(π/6)-2cos(2x-π/6)sin(π/6)-1
=2sin(2x-π/3)-1
x∈[-π/2,π/2]
2x-π/6∈[-4π/3,2π/3]
sin(2x-π/3)∈[-1,1]
y∈[-3,1]