1. Complex Number Representation: Let $x = \cos \theta + i \sin \theta$.
According to the properties of complex numbers:
$$\frac{1}{x} = \cos \theta - i \sin \theta$$
Adding these two expressions:
$$x + \frac{1}{x} = (\cos \theta + i \sin \theta) + (\cos \theta - i \sin \theta) = 2 \cos \theta$$
This matches our given condition.
2. Applying De Moivre's Theorem: De Moivre's Theorem states that $( \cos \theta + i \sin \theta )^n = \cos n\theta + i \sin n\theta$.
Therefore:
$$x^3 = (\cos \theta + i \sin \theta)^3 = \cos 3\theta + i \sin 3\theta$$
$$\frac{1}{x^3} = (\cos \theta - i \sin \theta)^3 = \cos 3\theta - i \sin 3\theta$$
3. Summation for $2 \cos 3\theta$: Now, add $x^3$ and $\frac{1}{x^3}$:
$$x^3 + \frac{1}{x^3} = (\cos 3\theta + i \sin 3\theta) + (\cos 3\theta - i \sin 3\theta)$$
$$x^3 + \frac{1}{x^3} = 2 \cos 3\theta$$
Thus, the value of $2 \cos 3\theta$ is $x^3 + \frac{1}{x^3}$.