f(x)=sin(2x+π/6)+sin(2x-π/6)+2cos²x=sin2xcosπ/6+sinπ/6cos2x+sin2xcosπ/6-sinπ/6cos2x+2cos²x=2cos²x+2sin2xcosπ/6=cos2x+1+√3sin2x=2sin(2x+π/6)+1f(x)的最大值是3最小值-1同样的题目（2）使f(x)大于等于2的x的取值范围是多少？刚才忘记求周期了T=2π/2=πf(x) =2sin(2x+π/6)+1≥2sin(2x+π/6)≥1/2π/6+2kπ≤2x+π/6≤5π/6+2kπkπ≤x≤π/3+kπ k∈z