The periods of sin(x+π/6) and sin(x-π/3) are both 2π; So take the image of sin2x as the reference of distribution control line.

In A and B, the mirror image of sin2x first increases and then decreases, which is the same as the mirror image of sinx on [0,2π]. So we can consider x=0 to judge the value of the starting point; At this time, sin (x+π/6) = sinπ/6 =1/2 >; 0; sin(x-π/3)=-sinπ/3 =-√3/2 & lt； 0; Is the case of a;

B is the starting point when x=π is considered; At this time, sin (x+π/6) = sin7π/6 =- 1/2.

Is the case of b;

C and D represent the situation that the images of sin2x are all reduced first and then increased; The starting points are located at x=π/2 and x= -π/2 respectively, which can be judged by the sign of the starting point value;

Of course, it can also be judged by monotonicity.