Ratio of Coefficients in Trigonometric Equation with Sum of Roots Given

To find the value of $\frac{b}{a}$, we start with the given trigonometric equation: $\sqrt{3} a \cos x + 2 b \sin x = c, \quad \text{where } x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

Since $\alpha$ and $\beta$ are two distinct real roots of this equation, they both satisfy it: $\sqrt{3} a \cos \alpha + 2 b \sin \alpha = c \quad \text{--- (1)}$ $\sqrt{3} a \cos \beta + 2 b \sin \beta = c \quad \text{--- (2)}$

Subtracting equation (2) from equation (1), we get: $\sqrt{3} a (\cos \alpha - \cos \beta) + 2 b (\sin \alpha - \sin \beta) = 0$

Using the sum-to-product identities: $\cos \alpha - \cos \beta = -2 \sin\left(\frac{\alpha + \beta}{2}\right) \sin\left(\frac{\alpha - \beta}{2}\right)$ $\sin \alpha - \sin \beta = 2 \cos\left(\frac{\alpha + \beta}{2}\right) \sin\left(\frac{\alpha - \beta}{2}\right)$

Substituting these identities into the equation gives: $\sqrt{3} a \left[-2 \sin\left(\frac{\alpha + \beta}{2}\right) \sin\left(\frac{\alpha - \beta}{2}\right)\right] + 2 b \left[2 \cos\left(\frac{\alpha + \beta}{2}\right) \sin\left(\frac{\alpha - \beta}{2}\right)\right] = 0$

Factoring out $2 \sin\left(\frac{\alpha - \beta}{2}\right)$: $2 \sin\left(\frac{\alpha - \beta}{2}\right) \left[ -\sqrt{3} a \sin\left(\frac{\alpha + \beta}{2}\right) + 2 b \cos\left(\frac{\alpha + \beta}{2}\right) \right] = 0$

Since $\alpha$ and $\beta$ are distinct roots ($\alpha \neq \beta$), we have $\frac{\alpha - \beta}{2} \neq 0$, which implies $\sin\left(\frac{\alpha - \beta}{2}\right) \neq 0$. Therefore, we can divide by this non-zero factor to obtain: $-\sqrt{3} a \sin\left(\frac{\alpha + \beta}{2}\right) + 2 b \cos\left(\frac{\alpha + \beta}{2}\right) = 0$

We are given that $\alpha + \beta = \frac{\pi}{3}$, which means $\frac{\alpha + \beta}{2} = \frac{\pi}{6}$. Substituting this into the equation: $-\sqrt{3} a \sin\left(\frac{\pi}{6}\right) + 2 b \cos\left(\frac{\pi}{6}\right) = 0$

Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$ and $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$: $-\sqrt{3} a \left(\frac{1}{2}\right) + 2 b \left(\frac{\sqrt{3}}{2}\right) = 0$

Dividing the entire equation by $\frac{\sqrt{3}}{2}$: $-a + 2b = 0 \implies 2b = a$

Thus, the ratio $\frac{b}{a}$ is: $\frac{b}{a} = \frac{1}{2} = 0.5$