f(x)=sin(2x+π/6)+2cosx^2-1
=sin(2x+π/6)+cos2x
=√3/2*sin2x+1/2*cos2x+cos2x
= √3/2*sin2x+3/2*cos2x
=√3*(1/2*sin2x+√3/2*cos2x)
=√3sin(2x+π/3)
单调递增区域为：
-π/2+2kπ≤2x+π/3≤π/2+2kπ,k为整数
-5π/6+2kπ≤2x≤π/6+2kπ,k为整数
-5π/12+kπ≤x≤π/12+kπ,k为整数
则函数f(x)的单调增区间为：[-5π/12+kπ,π/12+kπ],k为整数