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 need to find the value of \( \sqrt{3} \sin \alpha + \cos \alpha \) given the equation:

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

### Step 1: Use the cosine addition formula
We start by using the cosine addition formula:

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

Substituting this into the given equation:

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

This simplifies to:

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

### Step 2: Rearranging the equation
Rearranging gives us:

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

### Step 3: Use the identity for cosine and sine
We can use the identity \( \sin^2 x + \cos^2 x = 1 \) to express the equation in terms of a single variable. Let's denote \( \alpha = \beta \) (as a potential solution) to simplify our calculations:

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

Using the double angle formula \( \cos(2\alpha) = 2\cos^2(\alpha) - 1 \):

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

This simplifies to:

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

### Step 4: Rearranging further
Rearranging gives:

\[ -2\cos^2 \alpha + 2\cos \alpha + 1 - \frac{3}{2} = 0 \]

This leads to:

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

Multiplying through by -1 to simplify:

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

### Step 5: Solving the quadratic equation
Now we can multiply the entire equation by 2 to eliminate the fraction:

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

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

Here, \( a = 4, b = -4, c = 1 \):

\[ \cos \alpha = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 4 \cdot 1}}{2 \cdot 4} \]

Calculating the discriminant:

\[ \sqrt{16 - 16} = 0 \]

Thus:

\[ \cos \alpha = \frac{4}{8} = \frac{1}{2} \]

### Step 6: Finding sin alpha
Since \( \cos \alpha = \frac{1}{2} \), we can find \( \sin \alpha \):

Using \( \sin^2 \alpha + \cos^2 \alpha = 1 \):

\[ \sin^2 \alpha + \left(\frac{1}{2}\right)^2 = 1 \]

\[ \sin^2 \alpha + \frac{1}{4} = 1 \]

\[ \sin^2 \alpha = \frac{3}{4} \]

\[ \sin \alpha = \frac{\sqrt{3}}{2} \]

### Step 7: Substitute to find the final value
Now we substitute back to find \( \sqrt{3} \sin \alpha + \cos \alpha \):

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

Thus, the final answer is:

\[ \boxed{2} \]