Calculate the value of x, if x ∈ [-180°; 360°] cos2x = cosx + 2

$$2\cos^2 x - \cos x - 1 - 2 = 0$$ $$2\cos^2 x - \cos x - 3 = 0$$

Solve the quadratic equation $2y^2 - y - 3 = 0$.
Factor the quadratic expression:

$$(2y - 3)(y + 1) = 0$$

This gives two possible values for $y$:

$$2y - 3 = 0 \implies y = \frac{3}{2}$$ $$y + 1 = 0 \implies y = -1$$

So, we have $\cos x = \frac{3}{2}$ or $\cos x = -1$.

For $\cos x = \frac{3}{2}$:
There is no solution for this, as the range of the cosine function is $[-1, 1]$. The value $\frac{3}{2} = 1.5$ is outside this range.

For $\cos x = -1$:
The values of $x$ for which $\cos x = -1$ are $180^\circ, -180^\circ, 540^\circ, \dots$.
Considering the specified interval $x \in [-180^\circ; 360^\circ]$, the values are:

$$x = -180^\circ$$ $$x = 180^\circ$$

The values of $x$ in the given range are $\boxed{x = -180^\circ, 180^\circ}$.