3. From the meaning of the question: π/3-π/6 = (n+ 1/4) t, where t is the minimum positive period, T=2π/w, w = 3 (4n+1); If n is an integer, then w is an integer multiple of 9, depending on the meaning of the question, √ (1+a2) sin [3 (4n+1) π/3+u] = √ (1+a2) sin [(4n+65438+)] Or: from the title, f(π/3)=0 (*) and x=π/6 is the maximum point, so (π/3 -π/6)=|K+ 1/4|T, where k is an arbitrary integer and t is f(x). Therefore, w= 12K+3 and w >: substituting 0 into (*) gives: sin(π+4Kπ)+acos(π+4Kπ)=0. The simplification is as follows: sinπ+acosπ=0, so a=0, so a+w=| 12K+3|, and the option may take 3 or 9. In addition, because x = π/3 |.