Trigonometrical Equations 1 Question 21
21. Let $a, b, c$ be three non-zero real numbers such that the equation $\sqrt{3} a \cos x+2 b \sin x=c, x \in-\frac{\pi}{2}, \frac{\pi}{2}$, has two distinct real roots $\alpha$ and $\beta$ with $\alpha+\beta=\frac{\pi}{3}$. Then, the value of $\frac{b}{a}$ is
(2018 Adv.)
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Answer:
Correct Answer: 21. (0.5)
Solution:
- We have, $\alpha, \beta$ are the roots of
$ \begin{aligned} & \quad \sqrt{3} a \cos x+2 b \sin x=c \\ & \therefore \quad \sqrt{3} a \cos \alpha+2 b \sin \alpha=c \\ & \text { and } \quad \sqrt{3} a \cos \beta+2 b \sin \beta=c \end{aligned} $
On subtracting Eq. (ii) from Eq. (i), we get
$\sqrt{3} a(\cos \alpha-\cos \beta)+2 b(\sin \alpha-\sin \beta)=0$
$\Rightarrow \sqrt{3} a-2 \sin \frac{\alpha+\beta}{2} \quad \sin \frac{\alpha-\beta}{2}$
$ +2 b 2 \cos \frac{\alpha+\beta}{2} \quad \sin \frac{\alpha-\beta}{2}=0 $
$\Rightarrow \sqrt{3} a \sin \frac{\alpha+\beta}{2}=2 b \cos \frac{\alpha+\beta}{2}$
$\Rightarrow \tan \frac{\alpha+\beta}{2}=\frac{2 b}{\sqrt{3} a}$
$\Rightarrow \quad \tan \frac{\pi}{6}=\frac{2 b}{\sqrt{3} a} \quad \because \alpha+\beta=\frac{\pi}{3}$, given
$\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{2 b}{\sqrt{3} a} \Rightarrow \frac{b}{a}=\frac{1}{2}$
$\Rightarrow \quad \frac{b}{a}=0.5$
