Finding b/a,√3acos(x) + 2bsin(x) = c

Yup, it's another Saturday morning. I woke up knowing it was a free day. But Aah, when I picked up my phone to check the time, I saw a WhatsApp message asking for help in solving this:

Let a, b, and c be non-zero real numbers. The two distinct real solutions of the equation

are $\alpha$ and $\beta$ such that $\alpha + \beta = \frac{\pi}{3}$. Find the value of $\frac{b}{a}$. Here $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$.

Alright, let's start solving.

We start with

$\sqrt{3}a\cos x + 2b\sin x = c$

Since we need $\frac{b}{a}$, we can simply divide the entire equation by a, so we can get a head start.

$\sqrt{3}\cos x + \frac{2b}{a} \sin x = \frac{c}{a}$

And yes why not divide the whole equation by 2 as well?, Then we clearly have $\frac{b}{a}$ in the equation.

Let's do it quickly:

$\frac{\sqrt{3}}{2}\cos x + \frac{b}{a} \sin x = \frac{c}{2a}$

Now comes the fun part! Since there are two distinct solutions given, namely $\alpha$ and $\beta$ we can get two equations. Notice that if we subtract those two equations we can get ride of the R.H.S since it won't change.

Let's do that.

First substituting $x = \alpha$ :

$\frac{\sqrt{3}}{2}\cos \alpha + \frac{b}{a} \sin \alpha = \frac{c}{2a} \;\; \longrightarrow (1)$

Next substituting $x = \beta$ :

$\frac{\sqrt{3}}{2}\cos \beta + \frac{b}{a} \sin \beta = \frac{c}{2a} \;\; \longrightarrow (2)$

Now as we discussed earlier we just subtract, I'll do $(1) - (2)$ So R.H.S = 0 I will group this at once since it's easier and it's useless to write the entire subtraction.

$\frac{\sqrt{3}}{2} (\cos \alpha - \cos \beta) + \frac{b}{a} (\sin \alpha - \sin \beta) = 0$

Now, by using some basic trigonometric identities, we can simple transform these two expressions into a form that allows us to use the value of $\alpha + \beta$

Alright, that's a bit long, but wait, did you see it yet? They all going to cancel out but first, let's substitute the value for $\alpha + \beta$ .

Given $\alpha +\beta = \frac{\pi}{3}$

$\Rightarrow \frac{\alpha+\beta}{2} = \frac{\pi}{6}$

$\sin \frac{\alpha + \beta}{2} = \frac{1}{2}$ and $\cos \frac{\alpha + \beta}{2} = \frac{\sqrt{3}}{2}$

Now start cancelling out, the $\sqrt{3}$ 's will cancel out, $\sin\left(\frac{\alpha - \beta}{2}\right)$ will also cancel out.

Now we are left with a simple equation like this:

$\frac{-1}{2} + \frac{b}{a} = 0$

Yup done.

$\frac{b}{a} = \frac{1}{2}$

Whoever reads good luck!