已知√3sina／2＋[sin(a-π/2)]／[cos(π＋a／2)]*cosa／2＝1,a属于（0,2π）,求a的值.

3sin(a/2)＋[sin(a-π/2)]／[cos(π＋a/2)]*cos(a/2)＝√3sin(a/2)＋[-cosa]／[-cos(a/2)]*cos(a/2)＝√3sin(a/2)＋cosa＝1.
代入cosa＝1-2[sin(a/2)]^2,则√3sin(a/2)-2[sin(a/2)]^2＝0,所以sin(a/2)＝0或√3/2.
a∈(0,2π),则a/2∈(0,π),所以sin(a/2)=0无解,sin(a/2)＝√3/2的解是a/2=π/3或2π/3,即a=2π/3或4π/3.
所以,a=2π/3或4π/3.