1. The problem involves solving the equation $\cos x = \frac{1}{2}$ and $\cos x = -\frac{1}{2}$ for all possible values of $x$. 2. Recall that cosine is positive in the first and fourth quadrants, and negative in the second and third quadrants. 3. For $\cos x = \frac{1}{2}$, the principal solutions are: - $x_1 = \frac{\pi}{3} + 2k\pi$ (first quadrant) - $x_2 = -\frac{\pi}{3} + 2k\pi$ (equivalent to $\frac{5\pi}{3} + 2k\pi$, fourth quadrant) Here, $k$ is any integer representing the periodicity of cosine with period $2\pi$. 4. For $\cos x = -\frac{1}{2}$, the solutions are: - $x_3 = \frac{2\pi}{3} + 2k\pi$ (second quadrant) - $x_4 = -\frac{2\pi}{3} + 2k\pi$ (equivalent to $\frac{4\pi}{3} + 2k\pi$, third quadrant) 5. The expressions $x_1, x_2, x_3, x_4$ represent all angles where cosine takes the values $\frac{1}{2}$ and $-\frac{1}{2}$ respectively, repeated every $2\pi$ radians due to the periodic nature of cosine. 6. The integer $k \in \mathbb{Z}$ accounts for all such repetitions along the real number line. 7. The image's left group corresponds to the positive cosine solutions, and the right group corresponds to the negative cosine solutions, each with two symmetric angles about the x-axis. This explains the given equations and their geometric meaning on the unit circle.