Question

已知函数f(x)=根号3sin(2x-派/6)+2sin^2(x-派/12)，(x属于r)求函数f(x)的最小正周期

求使函数f(x)取得最大值的x的集合

Answer 1

f(x)=√3sin(2x-π/6)+2sin²(x-π/12)=√3sin(2x-π/6)-cos(2x-π/6)+1
=2sin(2x-π/3)+1
最小正周期为T=2π/ω=π,令2x-π/3=π/2+kπ（k为整数）,f(x)取得最大值的x的集合为x=5π/12+kπ/2（k为整数）

Answer 2

f(x)=√3sin(2x-π/6)+2sin²(x-π/12)
=√3sin(2x-π/6)+2[1-cos(2x-π/6)]/2
=√3sin(2x-π/6)-cos(2x-π/6)+1
=2sin(2x-π/6)+1
∵ω=2
∴最小正周期为T=2π/ω=2π/2=π

Answer 3

化简为f(x)=2sin(2x-π/3)+1，T最小=π，f(x)取得最大值的x的集合为x=5π/12+kπ/2（k为整数）