Let `a ,\ b ,\ c` be three non-zero real numbers such that the equation `sqrt(3)\ acosx+2\ bsinx=c ,\ x in [-pi/2,pi/2]` , has two distinct real roots `alpha` and `beta` with `alpha+beta=pi/3` . Then, the value of `b/a` is _______.

Explanation

To solve the problem, we start with the equation given:

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

We know that this equation has two distinct real roots \(\alpha\) and \(\beta\) such that:

\[ \alpha + \beta = \frac{\pi}{3} \]

### Step 1: Substitute the roots into the equation

Since \(\alpha\) and \(\beta\) are roots of the equation, we can write:

\[ \sqrt{3} a \cos \alpha + 2b \sin \alpha = c \quad \text{(1)} \]

\[ \sqrt{3} a \cos \beta + 2b \sin \beta = c \quad \text{(2)} \]

### Step 2: Set the two equations equal to each other

From equations (1) and (2), we can equate the right-hand sides:

\[ \sqrt{3} a \cos \alpha + 2b \sin \alpha = \sqrt{3} a \cos \beta + 2b \sin \beta \]

### Step 3: Rearranging the equation

Rearranging gives us:

\[ \sqrt{3} a \cos \alpha - \sqrt{3} a \cos \beta = 2b \sin \beta - 2b \sin \alpha \]

Factoring both sides:

\[ \sqrt{3} a (\cos \alpha - \cos \beta) = 2b (\sin \beta - \sin \alpha) \]

### Step 4: Use the sine and cosine difference identities

Using the identities for the difference of cosines and sines, we have:

\[ \cos \alpha - \cos \beta = -2 \sin\left(\frac{\alpha + \beta}{2}\right) \sin\left(\frac{\alpha - \beta}{2}\right) \]

\[ \sin \beta - \sin \alpha = 2 \cos\left(\frac{\alpha + \beta}{2}\right) \sin\left(\frac{\beta - \alpha}{2}\right) \]

Substituting these into our equation gives:

\[ \sqrt{3} a (-2 \sin\left(\frac{\pi/3}{2}\right) \sin\left(\frac{\alpha - \beta}{2}\right)) = 2b (2 \cos\left(\frac{\pi/3}{2}\right) \sin\left(\frac{\beta - \alpha}{2}\right)) \]

### Step 5: Simplifying the equation

Since \(\alpha + \beta = \frac{\pi}{3}\), we can simplify:

\[ \sqrt{3} a (-2 \sin\left(\frac{\pi}{6}\right) \sin\left(\frac{\alpha - \beta}{2}\right)) = 2b (2 \cos\left(\frac{\pi}{6}\right) \sin\left(\frac{\beta - \alpha}{2}\right)) \]

Knowing that \(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\) and \(\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\), we substitute these values:

\[ \sqrt{3} a (-2 \cdot \frac{1}{2} \sin\left(\frac{\alpha - \beta}{2}\right)) = 2b (2 \cdot \frac{\sqrt{3}}{2} \sin\left(\frac{\beta - \alpha}{2}\right)) \]

This simplifies to:

\[ -\sqrt{3} a \sin\left(\frac{\alpha - \beta}{2}\right) = 2b \sqrt{3} \sin\left(\frac{\beta - \alpha}{2}\right) \]

### Step 6: Canceling terms and solving for \(\frac{b}{a}\)

Since \(\sin\left(\frac{\alpha - \beta}{2}\right) = -\sin\left(\frac{\beta - \alpha}{2}\right)\), we can cancel out \(\sqrt{3}\) from both sides:

\[ -a = 2b \]

Thus, we find:

\[ \frac{b}{a} = -\frac{1}{2} \]

### Final Answer

The value of \(\frac{b}{a}\) is:

\[ \frac{b}{a} = -\frac{1}{2} \]