Question

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

Answer 1

So here in this question First we will assume $\tan \dfrac{x}{2}=t$, and we know two basic trigonometric formulas
$\cos x=\dfrac{1-{{\tan }^{2}}\dfrac{x}{2}}{1+{{\tan }^{2}}\dfrac{x}{2}}$ and similarly, $\sin x=\dfrac{2\tan \dfrac{x}{2}}{1+{{\tan }^{2}}\dfrac{x}{2}}$
So, we can write $\sqrt{3}a\cos x+2b\sin x=c$ as $\sqrt{3}a\dfrac{1-{{\operatorname{t}}^{2}}}{1+{{\operatorname{t}}^{2}}}+2b\dfrac{2\operatorname{t}}{1+{{\operatorname{t}}^{2}}}=c$
Which on solving equals to ${{\operatorname{t}}^{2}}(c+\sqrt{3}a)-4bt+c-\sqrt{3}a=0$
Whose two roots are $\tan \dfrac{\alpha }{2},\tan \dfrac{\beta }{2}$
Now applying formula $\tan (\dfrac{\alpha +\beta }{2})=\dfrac{\tan \dfrac{\alpha }{2}+\tan \dfrac{\beta }{2}}{1-\tan \dfrac{\alpha }{2}\tan \dfrac{\beta }{2}}$ we get desired result

Complete step-by-step solution:
Given an equation $\sqrt{3}a\cos x+2b\sin x=c$ which has roots $\alpha$ and $\beta$ with $\alpha +\beta =\dfrac{\pi }{3}$.
To find value of $\dfrac{2b}{a}$, first of all, we have to use formula $\cos x=\dfrac{1-{{\tan }^{2}}\dfrac{x}{2}}{1+{{\tan }^{2}}\dfrac{x}{2}}$ and similarly $\sin x=\dfrac{2\tan \dfrac{x}{2}}{1+{{\tan }^{2}}\dfrac{x}{2}}$
And convert equation $\sqrt{3}a\cos x+2b\sin x=c$ to $\sqrt{3}a\dfrac{1-{{\tan }^{2}}\dfrac{x}{2}}{1+{{\tan }^{2}}\dfrac{x}{2}}+2b\dfrac{2\tan \dfrac{x}{2}}{1+{{\tan }^{2}}\dfrac{x}{2}}=c$
Now assume $\tan \dfrac{x}{2}=t$ and now replacing it in equation it will look like
$\sqrt{3}a\dfrac{1-{{\operatorname{t}}^{2}}}{1+{{\operatorname{t}}^{2}}}+2b\dfrac{2\operatorname{t}}{1+{{\operatorname{t}}^{2}}}=c$ which on solving looks like ${{\operatorname{t}}^{2}}(c+\sqrt{3}a)-4bt+c-\sqrt{3}a=0$
Now this is a quadratic in t ,it was given that it x has two roots $\alpha$ and $\beta$, so t has two roots $\tan \dfrac{\alpha }{2},\tan \dfrac{\beta }{2}$
Using property for quadratic equation we can say ${{x}^{2}}(a)+bx+c=0$
Sum of roots equals to $-\dfrac{b}{a}$ and product of roots equals to $\dfrac{c}{a}$, on applying
Sum of roots $\tan \dfrac{\alpha }{2}+\tan \dfrac{\beta }{2}$ equals to $-\dfrac{b}{a}$ and product $\tan \dfrac{\alpha }{2}\times \tan \dfrac{\beta }{2}$ equals to $\dfrac{c}{a}$
$\tan \dfrac{\alpha }{2}+\tan \dfrac{\beta }{2}=\dfrac{4b}{c+\sqrt{3}a}and\tan \dfrac{\alpha }{2}\times \tan \dfrac{\beta }{2}=\dfrac{c-\sqrt{3}a}{c+\sqrt{3}a}...(2)$
Now it is given $\alpha +\beta =\dfrac{\pi }{3}$, dividing by 2 both side we $\dfrac{\alpha +\beta }{2}=\dfrac{\pi }{6}$
Now we can write it as $\tan (\dfrac{\alpha +\beta }{2})=\tan \dfrac{\pi }{6}$, now using formula $\tan (\dfrac{\alpha +\beta }{2})=\dfrac{\tan \dfrac{\alpha }{2}+\tan \dfrac{\beta }{2}}{1-\tan \dfrac{\alpha }{2}\tan \dfrac{\beta }{2}}$ and using equation (2)
We can write as
$\dfrac{\dfrac{4b}{c+\sqrt{3}a}}{1-\dfrac{c-\sqrt{3}a}{c+\sqrt{3}a}}=\tan \dfrac{\pi }{6}=\dfrac{1}{\sqrt{3}}$ , on further solving gives $\dfrac{4b}{c+\sqrt{3}a-(c-\sqrt{3}a)}=\tan \dfrac{\pi }{6}=\dfrac{1}{\sqrt{3}}$
After solving we get $\dfrac{4b}{2\sqrt{3}a}=\dfrac{1}{\sqrt{3}}$ which results into $\dfrac{2b}{a}=1$
Hence answer is $\dfrac{2b}{a}=1$.

Note: Some students have doubt that how to think that whether half angle formula to apply or not , look in this question $\sqrt{3}a\cos x+2b\sin x=c$ we have two roots $\alpha$ and $\beta$
We have two different trigonometry’s $\cos x,\sin x$ so to convert the whole equation into one format we use half angle formula and convert into $\tan \dfrac{x}{2}$.