x∈(π/12,π/3)
所以2x+π/3∈（π/2,π）
所以sin（2x+π/3）=12/13
sin2x=[(2x+π/3)-π/3]
=sin（2x+π/3）×cosπ/3-cos（2x+π/3）×sinπ/3
=12/13×(1/2)-(-5/13)×(√3/2)
=(12+5√3)/26