The value of $$\cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{4\pi}{7}\right) + \cos\left(\frac{6\pi}{7}\right)$$ is equal to

We need to find the value of $$\cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{4\pi}{7}\right) + \cos\left(\frac{6\pi}{7}\right)$$.

Using the roots of unity approach:

The 7th roots of unity are the solutions of $$z^7 = 1$$, given by $$z = e^{i \cdot 2k\pi/7}$$ for $$k = 0, 1, 2, \ldots, 6$$.

The sum of all 7th roots of unity is zero:

$$\sum_{k=0}^{6} e^{i \cdot 2k\pi/7} = 0$$

Taking the real part of both sides:

$$\sum_{k=0}^{6} \cos\left(\frac{2k\pi}{7}\right) = 0$$

$$1 + \cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{4\pi}{7}\right) + \cos\left(\frac{6\pi}{7}\right) + \cos\left(\frac{8\pi}{7}\right) + \cos\left(\frac{10\pi}{7}\right) + \cos\left(\frac{12\pi}{7}\right) = 0$$

Using the property $$\cos(2\pi - \theta) = \cos\theta$$:

$$\cos\left(\frac{12\pi}{7}\right) = \cos\left(2\pi - \frac{2\pi}{7}\right) = \cos\left(\frac{2\pi}{7}\right)$$

$$\cos\left(\frac{10\pi}{7}\right) = \cos\left(2\pi - \frac{4\pi}{7}\right) = \cos\left(\frac{4\pi}{7}\right)$$

$$\cos\left(\frac{8\pi}{7}\right) = \cos\left(2\pi - \frac{6\pi}{7}\right) = \cos\left(\frac{6\pi}{7}\right)$$

Substituting back:

$$1 + 2\left[\cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{4\pi}{7}\right) + \cos\left(\frac{6\pi}{7}\right)\right] = 0$$

Let $$S = \cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{4\pi}{7}\right) + \cos\left(\frac{6\pi}{7}\right)$$

$$1 + 2S = 0$$

$$S = -\frac{1}{2}$$

The correct answer is Option B: $$-\dfrac{1}{2}$$.