x∈[0,π]∴x+π/3∈[π/3,4π/3]画出大致图像x+π/3∈[π/3,2π/3]时有两个解sinπ/3=sin2π/3=√3/2sinπ/2=1∴√3/2<=m/2<1∴√3<=m<2m取值范围是[√3,2)两根关于x+π/3=π/2对称∴x1+π/3+x2+π...