If `0 lt alpha, beta lt pi` and they satisfy `cos alpha + cosbeta - cos(alpha + beta)=3/2`

Options

A. `alpha= beta`
B. `alpha + beta = (2pi)/3`
C. `alpha = 2beta`
D. `beta = 2alpha`

Answer

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

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

### Step 1: Rewrite the equation using the cosine addition formula
We know that:

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

However, we will use another identity for simplification:

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

Thus, we can rewrite the equation as:

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

### Step 2: Substitute the identity for \(\cos(\alpha + \beta)\)
Using the identity for \(\cos(\alpha + \beta)\):

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

Substituting this into our equation gives:

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

### Step 3: Simplify the equation
Rearranging yields:

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

This simplifies to:

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

### Step 4: Set up a quadratic equation
Let \(x = \cos\left(\frac{\alpha + \beta}{2}\right)\) and \(y = \cos\left(\frac{\alpha - \beta}{2}\right)\):

\[ 2xy - 2x^2 = 1 \]

Rearranging gives:

\[ 2x^2 - 2xy + 1 = 0 \]

### Step 5: Use the quadratic formula
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

Here, \(a = 2\), \(b = -2y\), and \(c = 1\):

\[ x = \frac{2y \pm \sqrt{(-2y)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} \]
\[ x = \frac{2y \pm \sqrt{4y^2 - 8}}{4} \]
\[ x = \frac{y \pm \sqrt{y^2 - 2}}{2} \]

### Step 6: Analyze the conditions
For \(x\) to be valid, \(y^2 - 2 \geq 0\) must hold, which implies \(y^2 \geq 2\). However, since \(y = \cos\left(\frac{\alpha - \beta}{2}\right)\), the maximum value of \(y\) is 1, indicating that \(y\) must equal 1.

### Step 7: Conclude with the values of \(\alpha\) and \(\beta\)
If \(y = 1\), then \(\frac{\alpha - \beta}{2} = 0\) leading to \(\alpha = \beta\).

Substituting back, we find:

\[ \alpha + \beta = 2\alpha = 2\cdot\frac{\pi}{3} \Rightarrow \alpha = \beta = \frac{\pi}{3} \]

Thus, we conclude:

1. \(\alpha = \beta\)
2. \(\alpha + \beta = \frac{2\pi}{3}\)

### Final Result
The correct options are:
- \(\alpha = \beta\)
- \(\alpha + \beta = \frac{2\pi}{3}\)