Let A_1 A_2 A_3 A_4 A_5 A_6 A_1 represent a regular hexagon, as...

Let the side length of the regular hexagon be $a$. The six sides of the hexagon taken in sequence can be represented as six vectors: $\overrightarrow{A_1 A_2}, \overrightarrow{A_2 A_3}, \overrightarrow{A_3 A_4}, \overrightarrow{A_4 A_5}, \overrightarrow{A_5 A_6}$, and $\overrightarrow{A_6 A_1}$.

Since these vectors form a closed polygon, their vector sum is zero:
$\overrightarrow{A_1 A_2} + \overrightarrow{A_2 A_3} + \overrightarrow{A_3 A_4} + \overrightarrow{A_4 A_5} + \overrightarrow{A_5 A_6} + \overrightarrow{A_6 A_1} = \vec{0}$

Consequently, the sum of their components along any axis must also be zero. Let us consider the components along the X-axis.

The angle made by each vector with the positive X-axis is as follows:
$\overrightarrow{A_1 A_2}$ is along the X-axis, so its angle is $0$.
$\overrightarrow{A_2 A_3}$ makes an angle of $60^{{}^{{}^{\circ}}}$ (or $\dfrac{\pi}{3}$) with the positive X-axis.
$\overrightarrow{A_3 A_4}$ makes an angle of $120^{{}^{{}^{\circ}}}$ (or $\dfrac{2\pi}{3}$) with the positive X-axis.
$\overrightarrow{A_4 A_5}$ makes an angle of $180^{{}^{{}^{\circ}}}$ (or $\dfrac{3\pi}{3}$) with the positive X-axis.
$\overrightarrow{A_5 A_6}$ makes an angle of $240^{{}^{{}^{\circ}}}$ (or $\dfrac{4\pi}{3}$) with the positive X-axis.
$\overrightarrow{A_6 A_1}$ makes an angle of $300^{{}^{{}^{\circ}}}$ (or $\dfrac{5\pi}{3}$) with the positive X-axis.

The sum of the x-components of all these vectors is zero:
$a \cos 0 + a \cos \dfrac{\pi}{3} + a \cos \dfrac{2\pi}{3} + a \cos \dfrac{3\pi}{3} + a \cos \dfrac{4\pi}{3} + a \cos \dfrac{5\pi}{3} = 0$

Dividing the entire equation by the magnitude $a$ ($a \neq 0$), we get:
$\cos 0 + \cos \dfrac{\pi}{3} + \cos \dfrac{2\pi}{3} + \cos \dfrac{3\pi}{3} + \cos \dfrac{4\pi}{3} + \cos \dfrac{5\pi}{3} = 0$