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

Hint: 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 \[\tan \dfrac{x}{2}\].