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(C) Given the equation $\sqrt{3} a \cos x + 2 b \sin x = c$.Dividing by $a$,we get $\sqrt{3} \cos x + \frac{2b}{a} \sin x = \frac{c}{a}$.Since $\alpha

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$0.5$

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(C) Given the equation $\sqrt{3} a \cos x + 2 b \sin x = c$.Dividing by $a$,we get $\sqrt{3} \cos x + \frac{2b}{a} \sin x = \frac{c}{a}$.Since $\alpha$ and $\beta$ are roots,they satisfy the equation:$\sqrt{3} \cos \alpha + \frac{2b}{a} \sin \alpha = \frac{c}{a} \quad (1)$$\sqrt{3} \cos \beta + \frac{2b}{a} \sin \beta = \frac{c}{a} \quad (2)$Subtracting $(2)$ from $(1)$:$\sqrt{3}(\cos \alpha - \cos \beta) + \frac{2b}{a}(\sin \alpha - \sin \beta) = 0$Using the sum-to-product formulas:$\sqrt{3} \left( -2 \sin \frac{\alpha + \beta}{2} \sin \frac{\alpha - \beta}{2} ight) + \frac{2b}{a} \left( 2 \cos \frac{\alpha + \beta}{2} \sin \frac{\alpha - \beta}{2} ight) = 0$Given $\alpha + \beta = \frac{\pi}{3}$,so $\frac{\alpha + \beta}{2} = \frac{\pi}{6}$.Since $\alpha 
eq \beta$,$\sin \frac{\alpha - \beta}{2} 
eq 0$,we can divide by it:$-\sqrt{3} \sin \frac{\pi}{6} + \frac{2b}{a} \cos \frac{\pi}{6} = 0$$-\sqrt{3} \left( \frac{1}{2} ight) + \frac{2b}{a} \left( \frac{\sqrt{3}}{2} ight) = 0$$-\frac{\sqrt{3}}{2} + \frac{b}{a} \sqrt{3} = 0$$\frac{b}{a} \sqrt{3} = \frac{\sqrt{3}}{2} \implies \frac{b}{a} = \frac{1}{2} = 0.5$.

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

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