Let a, b, c be three non-zero real numbers such that the equation √3acosx + 2bsinx = c, x∈[-π/2, π/2] has two distinct real roots α and β with α + β = π/3. Then the value of 2b/a is ___________.

√3acosx + 2bsinx = c
⇒ √3cosα + 2bsinα = c ..(1)
⇒ √3cosβ + 2bsinβ = c ..(2)
Consider (1) - (2)
⇒ √3[cosα - cosβ] + 2b(sinα - sinβ) = 0
⇒ √3[-2sin((α+β)/2)sin((α-β)/2)] + 2b[2cos((α+β)/2)sin((α-β)/2)] = 0
Since α + β = π/3, we have
⇒ -√3[2sin(π/6)sin((α-β)/2)] + 4b[cos(π/6)sin((α-β)/2)] = 0
⇒ -√3sin((α-β)/2) + 2√3b sin((α-β)/2) = 0
Since α and β are distinct, sin((α-β)/2) ≠ 0
⇒ -√3 + 2√3b = 0
⇒ 2√3b = √3
⇒ b = 1/2
Therefore, 2b/a = 1