sinx+根号3cosx+a=0在x∈[0,π/2]上有两个不同的实根,求实数a的取值范围

sinx+根号下3cosx+a=0
sinx+根号下3cosx=2sin(x+π/3)
x∈[0,π/2] x+π/3∈[π/3,5π/6]
2sin(x+π/3)∈[1,2]
因为有两个根 所以x+π/3∈[π/3,2π/3]
2sin(x+π/3)∈[根号下3,2]
所以a∈[-2,-根号下3]
即 -2