Discussion Forum

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

Answer:

Explanation:

We have $\alpha$ and $\beta$  are the roots of

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

$\therefore$    $\sqrt{3}a \cos \alpha +2b \sin \alpha=c$     .....(i)

 and       $\sqrt{3}a \cos \beta +2b \sin \beta=c$      .......(ii)

On substracting Eq. (ii) from Eq. (i) , we get

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

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

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

 $\Rightarrow$  $\tan(\frac{\alpha+\beta}{2})=\frac{2b}{\sqrt{3}a}$

$\Rightarrow$  $\tan(\frac{\pi}{6})=\frac{2b}{\sqrt{3}a}$      [ $\therefore$      $\alpha$ + $\beta$= $\frac{\pi}{3}$, given ]

$\Rightarrow$   $\frac{1}{\sqrt{3}}=\frac{2b}{\sqrt{3}a}\Rightarrow \frac{b}{a}=\frac{1}{2}$

  $\Rightarrow$     $\frac{b}{a}=0.5$