Quadratic equation in trigonometric form $\sqrt 3 a\cos x + 2b\sin x = c$ Let a,b,c be three non-zero real number such that the equation, $\sqrt 3 a\cos x + 2b\sin x = c$, $x \in \left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$, has two distict roots $\alpha$ and $\beta $ where $\alpha +\beta=\frac{\pi}{6}$. Then find the value of $\frac{b}{a}$.
My approach is as follow
$\sqrt 3 a\cos x + 2b\sin x = c$
$3{a^2}{\cos ^2}x = {c^2} + 4{b^2}{\sin ^2}x - 4bc\sin x$
$ {\sin ^2}x\left( {4{b^2} + 3{a^2}} \right) - 4bc\sin x + {c^2} - 3{a^2} = 0$
Roots are real when $16{b^2}{c^2} + 4\left( {4{b^2} + 3{a^2}} \right)\left( {{c^2} - 3{a^2}} \right) > 0$
$\sin \alpha  + \sin \beta  = \frac{{4bc}}{{4{b^2} + 3{a^2}}};\sin \alpha \sin \beta  = \frac{{{c^2} - 3{a^2}}}{{4{b^2} + 3{a^2}}}$, how I will proceed from here

let $\alpha =A,\beta =B$
as $$c-c=0$$ $$\sqrt{3}a \cos A+2b\sin A-\sqrt{3}a\cos B-2b\sin B=0$$
$$\sqrt{3}a \cos A-\sqrt{3}a\cos B+ 2b\sin A-2b\sin B=0$$
$$-2\sqrt{3}a \sin\left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)+2b\sin \left(\frac{A-B}{2}\right)\cos \left(\frac{A+B}{2}\right)=0$$  Can you end it now? :cancel $\sin (\frac{A-B}{2})$
