Quadratic equation in trigonometric form $\sqrt 3 a\cos x + 2b\sin x = c$

Question

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

Answer 1

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 $\sin (\frac{A-B}{2})$

Answer 2

You can see the equation as defining the intersections of the right half of the unit circle and a straight line. The bisector of the two points makes the angle $\dfrac{\alpha+\beta}2$ with the horizontal axis, and the line is perpendicular to it.

Hence

$$\dfrac{2b}{\sqrt3 a}=\tan\dfrac\pi{12}.$$

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Analytical solution:

$$\cos\left(x-\arctan\dfrac{2b}{\sqrt3a}\right)=\frac c{\sqrt{3a^2+4b^2}}$$

and

$$x=\arctan\dfrac{2b}{\sqrt3a}\pm\arccos\frac c{\sqrt{3a^2+4b^2}}.$$

Obviously,

$$\alpha+\beta=2\arctan\dfrac{2b}{\sqrt3a}.$$