求函数f(x)=√3sin(2x-π/6)+2平方sin(x-π/12)的最小正周期及取得最大值x的集合?有急用!
问题描述:
求函数f(x)=√3sin(2x-π/6)+2平方sin(x-π/12)的最小正周期及取得最大值x的集合?
有急用!
答
f(x)=√3sin(2x-π/6)+2sin²(x-π/12)=√3sin(2x-π/6)-cos(2x-π/6)+1=2sin(2x-π/3)+1.
故T=π,由2x-π/3=2kπ+π/2得,x=kπ+5π/12,k∈Z,
函数f(x)取得最大值x的集合{x|x=x=kπ+5π/12,k∈Z}
答
f(x)=√3sin(2x-π/6)+2平方sin(x-π/12)=√3sin(2x-π/6)+1-cos(2x-π/6)=2(√3/2sin(2x-π/6)-1/2cos(2x-π/6))+1=2(sin(2x-π/6)cosπ/6-cos(2x-π/6)sinπ/6)+1=2sin(2x-π/6-π/6)+1=2sin(2x-π/3)...