三角函数提问.

sinx＋√3cosx
=2(1/2sinx+√3/2cosx)
=2(sinxcosπ/3+cosxsinπ/3)
=2sin(x+π/3)
所以sin(x+π/3)=-a/2

0<x<π
π/3<x+π/3<4π/3
当2π/3<=x+π/3<4π/3
sin(x+π/3)是减函数，最大是√3/2
而π/3<x<2π/3.√3/2<sin(x+π/3)<=1
所以若2π/3<=x+π/3<4π/3，只有一个根
所以两个根则π/3<x+π/3+2π/3
且x+π/3≠π/2，因为此时也是一个解
所以√3/2<-a/2<1
-2<a<-√3

sin(x+π/3)=-a/2有两个根
则x+π/3关于π/2对称
所以(b+π/3)+(c+π/3)=2×π/2
b+c=π/3

-2<a<-根号3 b+c=π/3