首先,你的题目可能错了,中间应有一个系数是2.最后一项的系数也应有个2;
(1)f (x)＝√3cos2ωx＋2sinωxcosωx＋2a=2sin(2ωx+π/3)＋2a
f (x)的图象在y轴右侧的第一个最高点的横坐标为π/6,得2ωxπ/6+π/3=π/2得ω=1/2
（2）f (x)=2sin(2ωx+π/3)＋2a得f (x)＝2sin(x+π/3)＋2a
在区间[－π/3,5π/6]得x+π/3在区间[0,7π/6]得f (x)的最小值=2sin(7π/6)＋2a=√3得 a=√3+1）/2