Question

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，谢谢

Answer 1

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]