Is there a closed-form of $\sum_{n=1}^{\infty }\frac{\cos^2(n)}{n^2}$ Is there a closed-form of $$\sum_{n=1}^{\infty }\frac{\cos^2(n)}{n^2}$$

Since over the interval $(0,\pi)$ we have:
$$\frac{\pi-x}{2}=\sum_{n\geq 1}\frac{\sin(nx)}{n}\tag{1}$$
over the same interval we have also:
$$\frac{2\pi x-x^2}{4} = \sum_{n\geq 1}\frac{1-\cos(nx)}{n^2}\tag{2} $$
or:
$$\frac{2\pi x-x^2}{8} = \sum_{n\geq 1}\frac{\sin^2\left(n\frac{x}{2}\right)}{n^2}\tag{3} $$
so by setting $x=2$ we get:
$$\sum_{n\geq 1}\frac{\sin^2 n}{n^2}=\frac{\pi-1}{2}, \qquad \sum_{n\geq 1}\frac{\cos^2 n}{n^2}=\color{red}{\frac{\pi^2-3\pi+3}{6}}.\tag{4}$$
