一道高中三角函数的数学题,

sin(α+π/3)=1/2sinα+(根号3)/2cosa=1/4cosa=±√[1-(sina)^2]sinα+(√3){±√[1-(sina)^2]}=1/2(sinα-1/2)^2=1-(sina)^22(sina)^2-sina-3/4=02[(sina)^2-1/2sina+1/16-1/16-3/8]=02[(sina)-1/4]^2-7/8=0sina=7/...