Let \(a, b, c\) be three non-zero real numbers such that the equation \(\sqrt{3} \mathrm{a} \cos x+2 b \sin x=c\), \(x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) has two distinct real roots \(\alpha\) and \(\beta\) with \(\alpha+\beta=\frac{\pi}{3}\). Then the value of \(\frac{b}{a}\) is

\(\alpha\) and \(\beta\) are roots of \(\sqrt{3} \mathrm{a} \cos x+2 \mathrm{~b} \sin x=\mathrm{c}\)
\(\begin{aligned}
& \therefore \quad \sqrt{3} a \cos \alpha+2 b \sin \alpha=c ...(i)\\
& \sqrt{3} a \cos \beta+2 b \sin \beta=c...(ii)
\end{aligned}\)
Subtracting (ii) from (i), we get
\(\begin{aligned}
& \sqrt{3} a(\cos \alpha-\cos \beta)+2 b(\sin \alpha-\sin \beta)=0 \\
& \begin{aligned}
& \Rightarrow \sqrt{3} a[ \left.-2 \sin \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\alpha-\beta}{2}\right)\right] \\
&++2 b\left[2 \cos \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\alpha-\beta}{2}\right)\right]=0 \\
& \Rightarrow-\sqrt{3} a\left[2 \sin \left(\frac{\pi}{6}\right) \sin \left(\frac{\alpha-\beta}{2}\right)\right] \\
&++2 b\left[2 \cos \left(\frac{\pi}{6}\right) \sin \left(\frac{\alpha-\beta}{2}\right)\right]=0
\end{aligned}
\end{aligned}\)
\(\begin{aligned} & \Rightarrow-\sqrt{3} a \sin \left(\frac{\alpha-\beta}{2}\right)+2 b\left[\sqrt{3} \sin \left(\frac{\alpha-\beta}{2}\right)\right]=0 \\ & \Rightarrow-\sqrt{3} a+2 \sqrt{3} b=0 \Rightarrow \frac{b}{a}=\frac{1}{2}=0.5\end{aligned}\)