A = \left\{ \cos (4k^2\pi) \cdot\displaystyle\frac{2-6k}{2k+1} :k \in\mathbb{N} \right\} \cup \left\{ \cos (4k^2\pi - 4k\pi +\pi) \cdot\displaystyle\frac{5-6k}{2k}\right\} \\ = \left\{ -3 + \displaystyle\frac{5}{2k+1} : k\in\mathbb{N} \right\} \cup \left\{ 3 - \displaystyle\frac{5}{2k} :k\in\mathbb{N} \right\} \implies \sup A = 3, \quad\inf A = -3