已知函数f(x)=sin(2x+φ)+αcos(2x+φ)其中αφ为正常数且0

(1)由f(x)=sin(2x+φ)+acos(2x+φ)≤√1+a²
则由√1+a²=2及a＞0,求得a=√3
于是f(x)=sin(2x+φ)+ √3cos(2x+φ)
=2〔sin(2x+φ)·cosπ/3+cos(2x+φ)·sinπ/3〕
=2sin(2x+π/3+φ).
又f(x)的图象关于直线x=π/6对称,则在x=π/6时,f(x)取最值.
则有2·π/6+φ+π/3=kπ+π/2,则φ=kπ+π/2-2π/3=kπ-π/6 (k∈Z).
又0＜φ＜π,求得φ=5π/6
(2)由(1)可知f(x)=2sin(2x+7π/6).
由y=f(x)的纵坐标不变,而横坐标在x轴方向向右平移5π/12个单位即可.