We know that, for any rational number $p$, we have that $\cos(p\pi)$ is an algebraic number.
Since this property comes from the fact that $e^{ip\pi}$ is algebraic (as a root of unity), I suspect that $\pi$ is the unique transcendental number with such property, in the sense that there does not exists another transcendental number $\alpha\ne q \pi$, for rational $q$, such that $\cos(p\alpha)$ is an algebraic number. But I don't find a proof. It is true?
This answer shows that $\alpha=\cos^{-1}(3/5)$ is not a rational multiple of $\pi$. Additionally, the Lindemann-Weierstrass theorem shows that $\alpha$ is transcendental.
Nevertheless, $\cos(p\alpha)$ is algebraic for all rational $p$. If $p=\frac{n}{m}$, then $\cos(p \alpha)$ satisfies the polynomial equation $T_m(\cos(p \alpha))=T_n(3/5)$, where $T_k$ is the $k$th Chebyshev polynomial.