`cos.(2pi)/(7)+cos.(4pi)/(7)+cos.(6pi)/(7)`

Options

A. is equal to zero
B. lies between 0 and 3
C. is a negative number
D. lies between 3 and 6

Explanation

To solve the expression \( \cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{4\pi}{7}\right) + \cos\left(\frac{6\pi}{7}\right) \), we can follow these steps:

### Step 1: Set Up the Equation
Let \( \theta = \frac{2\pi}{7} \). Then, we can rewrite the expression as:
\[ \cos(\theta) + \cos(2\theta) + \cos(3\theta) \]

### Step 2: Use the Cosine Addition Formula
We know that:
\[ \cos(3\theta) = 4\cos^3(\theta) - 3\cos(\theta) \]
We can substitute this into our expression:
\[ \cos(\theta) + \cos(2\theta) + (4\cos^3(\theta) - 3\cos(\theta)) \]
This simplifies to:
\[ 4\cos^3(\theta) - 2\cos(\theta) + \cos(2\theta) \]

### Step 3: Use the Double Angle Formula
Next, we can use the double angle formula for cosine:
\[ \cos(2\theta) = 2\cos^2(\theta) - 1 \]
Substituting this into our expression gives:
\[ 4\cos^3(\theta) - 2\cos(\theta) + (2\cos^2(\theta) - 1) \]
This simplifies to:
\[ 4\cos^3(\theta) + 2\cos^2(\theta) - 2\cos(\theta) - 1 \]

### Step 4: Set Up the Polynomial
Now, we can set up the polynomial:
\[ 4x^3 + 2x^2 - 2x - 1 = 0 \]
where \( x = \cos(\theta) \).

### Step 5: Find the Roots
The roots of this polynomial are \( \cos\left(\frac{2\pi}{7}\right) \), \( \cos\left(\frac{4\pi}{7}\right) \), and \( \cos\left(\frac{6\pi}{7}\right) \).

### Step 6: Use Vieta's Formulas
By Vieta's formulas, the sum of the roots of the polynomial \( ax^3 + bx^2 + cx + d = 0 \) is given by:
\[ -\frac{b}{a} \]
In our case:
\[ -\frac{2}{4} = -\frac{1}{2} \]

### Conclusion
Thus, we have:
\[ \cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{4\pi}{7}\right) + \cos\left(\frac{6\pi}{7}\right) = -\frac{1}{2} \]