The value of `2alpha+beta(0ltalpha, beta lt (pi)/(2))`, satisfying the equation `cos alpha cos beta cos (alpha+beta)=-(1)/(8)` is equal to

Options

A. `(5)/(6)pi`
B. `(pi)/(2)`
C. `pi`
D. `(7pi)/(12)`

Answer

To solve the problem, we need to find the value of \(2\alpha + \beta\) given the equation:

\[ \cos \alpha \cos \beta \cos(\alpha + \beta) = -\frac{1}{8} \]

where \(0 < \alpha, \beta < \frac{\pi}{2}\).

### Step 1: Rewrite the equation
We start by rewriting the equation. We can multiply both sides by 2:

\[ 2 \cos \alpha \cos \beta \cos(\alpha + \beta) = -\frac{1}{4} \]

### Step 2: Use the cosine addition formula
Using the cosine addition formula, we know:

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

Substituting this into our equation gives:

\[ 2 \left(\frac{1}{2}(\cos(\alpha + \beta) + \cos(\alpha - \beta))\right) \cos(\alpha + \beta) = -\frac{1}{4} \]

This simplifies to:

\[ (\cos(\alpha + \beta) + \cos(\alpha - \beta)) \cos(\alpha + \beta) = -\frac{1}{4} \]

### Step 3: Let \(t = \cos(\alpha + \beta)\)
Let \(t = \cos(\alpha + \beta)\). Then we have:

\[ t^2 + t \cos(\alpha - \beta) = -\frac{1}{4} \]

Rearranging gives:

\[ 4t^2 + 4t \cos(\alpha - \beta) + 1 = 0 \]

### Step 4: Use the discriminant condition
For the roots to be real, the discriminant must be non-negative:

\[ D = (4 \cos(\alpha - \beta))^2 - 4 \cdot 4 \cdot 1 \geq 0 \]

This simplifies to:

\[ 16 \cos^2(\alpha - \beta) - 16 \geq 0 \]

Dividing by 16 gives:

\[ \cos^2(\alpha - \beta) \geq 1 \]

This implies:

\[ \cos(\alpha - \beta) = 1 \implies \alpha - \beta = 0 \implies \alpha = \beta \]

### Step 5: Substitute \(\alpha = \beta\)
Substituting \(\alpha = \beta\) into our equation gives:

\[ 4t^2 + 4t \cdot 1 + 1 = 0 \]

This simplifies to:

\[ 4t^2 + 4t + 1 = 0 \]

### Step 6: Solve for \(t\)
Using the quadratic formula:

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

### Step 7: Find \(\alpha + \beta\)
Since \(t = \cos(\alpha + \beta)\), we have:

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

This means:

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

### Step 8: Find \(\alpha\) and \(\beta\)
Since \(\alpha = \beta\):

\[ 2\alpha = \frac{2\pi}{3} \implies \alpha = \frac{\pi}{3}, \beta = \frac{\pi}{3} \]

### Step 9: Calculate \(2\alpha + \beta\)
Now we can find \(2\alpha + \beta\):

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

Thus, the final answer is:

\[ \boxed{\pi} \]