If `0ltalpha,betaltpi` and `cos alpha+cos beta -cos(alpha+beta)=(3)/(2)`, then the value of `sqrt3 sin alpha+cos alpha` is equal to

Answer

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

\[ \cos \alpha + \cos \beta - \cos(\alpha + \beta) = \frac{3}{2} \]

### Step 1: Rearranging the Equation
We can rearrange the equation as follows:

\[ \cos \alpha + \cos \beta = \frac{3}{2} + \cos(\alpha + \beta) \]

### Step 2: Using the Cosine Addition Formula
Using the cosine addition formula, we know that:

\[ \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \]

Substituting this into our rearranged equation gives:

\[ \cos \alpha + \cos \beta = \frac{3}{2} + \cos \alpha \cos \beta - \sin \alpha \sin \beta \]

### Step 3: Multiplying by 2
To eliminate the fraction, we can multiply the entire equation by 2:

\[ 2(\cos \alpha + \cos \beta) = 3 + 2(\cos \alpha \cos \beta - \sin \alpha \sin \beta) \]

### Step 4: Rearranging Again
Rearranging gives us:

\[ 2\cos \alpha + 2\cos \beta - 2\cos \alpha \cos \beta + 2\sin \alpha \sin \beta = 3 \]

### Step 5: Using Trigonometric Identities
We can use the identity \( \sin^2 \theta + \cos^2 \theta = 1 \) to express the equation in terms of squares.

### Step 6: Setting Up a Quadratic Equation
We can express \( \cos \alpha + \cos \beta \) in terms of \( x = \cos \alpha \) and \( y = \cos \beta \):

\[ x + y - xy = \frac{3}{2} \]

### Step 7: Solving for \( \alpha \) and \( \beta \)
From the equation \( \sin \alpha = \sin \beta \), we conclude that \( \alpha = \beta \).

### Step 8: Finding the Value of \( \alpha \)
Substituting \( \alpha = \beta \) into the equation gives:

\[ 2\cos \alpha - \cos(2\alpha) = \frac{3}{2} \]

Using \( \cos(2\alpha) = 2\cos^2 \alpha - 1 \):

\[ 2\cos \alpha - (2\cos^2 \alpha - 1) = \frac{3}{2} \]

This simplifies to:

\[ 2\cos \alpha - 2\cos^2 \alpha + 1 = \frac{3}{2} \]

### Step 9: Solving the Quadratic
Rearranging gives:

\[ 2\cos^2 \alpha - 2\cos \alpha + \frac{1}{2} = 0 \]

Multiplying through by 2:

\[ 4\cos^2 \alpha - 4\cos \alpha + 1 = 0 \]

### Step 10: Using the Quadratic Formula
Using the quadratic formula \( \cos \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ \cos \alpha = \frac{4 \pm \sqrt{16 - 16}}{8} = \frac{4}{8} = \frac{1}{2} \]

### Step 11: Finding \( \alpha \)
Since \( \cos \alpha = \frac{1}{2} \), we find:

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

### Step 12: Finding \( \sqrt{3} \sin \alpha + \cos \alpha \)
Now we calculate:

\[ \sin \alpha = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}, \quad \cos \alpha = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

Thus,

\[ \sqrt{3} \sin \alpha + \cos \alpha = \sqrt{3} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{3}{2} + \frac{1}{2} = \frac{4}{2} = 2 \]

### Final Answer
The value of \( \sqrt{3} \sin \alpha + \cos \alpha \) is:

\[ \boxed{2} \]