f(x)=√3*sinxcosx+cos²x =√3/2*sin2x+1/2*(1+cos2x) =√3/2*sin2x+1/2*cos2x+1/2 =sin(2x+π/6)+1/2那么f'(x)=2cos(2x+π/6)≥0所以-π/2+2kπ≤2x+π/6≤π/2+2kπ于是-π/3+kπ≤x≤π/6+kπ即单调递增区间为:[-π/3+kπ,π/6+kπ] 结果...