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

$\text{Given: } \sqrt{3}\,a\cos x + 2b\sin x = c$

$x \in \left[-\dfrac{\pi}{2},\ \dfrac{\pi}{2}\right]$

$\text{two distinct real roots } \alpha,\ \beta, \quad \alpha + \beta = \dfrac{\pi}{3}$

$\text{To find: } \dfrac{b}{a}$

$1.\ \text{combine LHS: } P\cos x+Q\sin x=R\cos(x-\varphi)$ $2.\ \alpha+\beta=2\varphi \ \text{(symmetry about the peak)}$ $3.\ \text{use } \alpha+\beta=\dfrac{\pi}{3} \text{ to get } \varphi, \text{ then } \tan\varphi$ $P=\sqrt3\,a,\quad Q=2b$ $P\cos x + Q\sin x = c$ $P\cos x + Q\sin x = R\cos(x-\varphi)$ $R\cos(x-\varphi)\in[-R,\,R]$ $\text{two distinct roots} \iff |c|<R$ $R\cos(x-\varphi)=c \ \Rightarrow \ \cos(x-\varphi)=\dfrac{c}{R}$ $\text{cosine is symmetric about its peak } x=\varphi$ $\therefore\ \dfrac{\alpha+\beta}{2}=\varphi \ \Rightarrow \ \alpha+\beta=2\varphi$ $\alpha+\beta=\dfrac{\pi}{3} \ \Rightarrow \ 2\varphi=\dfrac{\pi}{3}$ $\varphi=\dfrac{\pi}{6}$ $R\cos(x-\varphi)=R\cos\varphi\cos x+R\sin\varphi\sin x$ $\Rightarrow\ R\cos\varphi=P,\quad R\sin\varphi=Q$ $R=\sqrt{P^2+Q^2},\qquad \tan\varphi=\dfrac{R\sin\varphi}{R\cos\varphi}=\dfrac{Q}{P}$ $\tan\varphi=\dfrac{Q}{P}=\dfrac{2b}{\sqrt3\,a}$ $\tan\dfrac{\pi}{6}=\dfrac{1}{\sqrt3}$ $\dfrac{1}{\sqrt3}=\dfrac{2b}{\sqrt3\,a}$ $2b=a$ $\dfrac{b}{a}=\dfrac{1}{2}$ $\boxed{\dfrac{b}{a}=\dfrac{1}{2}}$