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
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