此时，sin(2x+π/3)-1=a，由此可知a的取值范围为-1≤a≤1。
∵-π/6≤x≤13π/12∴0≤2x+π/3≤5π/2sin(2x+π/3)=a+1令t=2x+π/3∈[0,5π/2]sint=a+1（画出sint在[0,5π/2]的函数图像）a+1=1或-1≤a+1<0解得a=0或-2≤a<-1
