1 A=2 T=2π/2=π 初相=π/3 2 2x+π/3=π/2+kπ 所以对称轴为x=π/12+kπ/2 -π/2+2kπ≦2x+π/3≦π/2+2kπ 所以-5π/12+kπ≦x≦π/12+kπ 所以递增区间为[-5π/12+kπ,π/12+kπ ]