已知函数f(x)=根号3sin(2x减6分之派)加2sin平方(x减12分之派)(x属于R) 求函数

网友回答

f(x)=√3sin(2x-π/6）+2sin^2(x-π/12)
=√3sin(2x-π/6）+1-cos(2x-π/6）
=2sin(2x-π/6-π/6)+1
=2sin(2x-π/3）+1
当2x-π/3=2kπ+π/2,即x=kπ+5π/12（k∈Z）时f(x)取最大值
写成集合形式
======以下答案可供参考======
供参考答案1:
f(x)=√3sin(2x-π/6）+2sin^2(x-π/12)
=√3sin(2x-π/6）-cos(2x-π/6）+1
=2sin(2x-π/6-π/6)+1
=2sin(2x-π/3）+1
当2x-π/3=2kπ+π/2，
即{x|x=kπ+5π/12}（k∈Z）时f(x)取最大值
供参考答案2:
f(x)=√3sin(2x-pi/6)+2sin²(x-pi/12）
=√3sin(2x-pi/6)-[1-2sin²(x-pi/12）]+1
=√3sin(2x-pi/6)-cos(2x-pi/6)+1
=2sin(2x-pi/3)+1
2x-pi/3=pi/2+2k*pi(k属于整数) ==》｛x|x=5/12*pi+k*pi,k属于Z｝
供参考答案3:
f(x)=√3sin(2x-π/6)+1-cos(2x-π/6) =√3sin(2x-π/6)-cos(2x-π/6)+1
=2[√3/2sin(2x-π/6)-1/2cos(2x-π/6)]+1 =2[cos(π/6)sin(2x-π/6)-sin(π/6)cos(2x-π/6)]+1=2[sin[(2x-π/6)-π/6]+1 当(2x - π/6)-π/6 = 2kπ + π/2时，取得最大值 即解出x=kπ+5π/12 集合为{x|x=kπ+5π/12}