No.
As Henning said, $\cos^{-1} \alpha$ is a rational multiple of $\pi$ if and only if $z = \alpha+i\sqrt{1-\alpha^2}$ is a root of unity. Note that as a sum of algebraic numbers, $z$ is itself an algebraic number; and to find out for which $F \in \mathbb{Q}[x]$ is true that $F(z)=0$ we only need the resolvent $F(z) = \rho(q(y),p(z-x))$, where $q \in \mathbb{Q}[y]$ is the polynomial for which $q(\alpha)=0$, and $p\in\mathbb{Q}[x]$ is the polynomial for which $p(i\sqrt{1-\alpha^2})=0$.
But using this answer we can easily decide whether $z$ is a root of unity:
An algebraic $z$ of modulus one is a root of unity if and only if its (irreducible) $F$ is a cyclotomic polynomial. Let $N = \deg F$; there are finitely many cyclotomic polynomials $\Phi_M$ such that $\deg\Phi_M = \varphi(M) = N$, where $\varphi$ is Euler's totient function, so we only need to compare them.
For example, in the case of $z = 3/5 + i4/5$, $F(x) = x^2-\frac{6}{5}x+1$, which can never be a cyclotomic polynomial, since they have only integer coefficients, as roots of unity are algebraic integers.
In the case of $\varphi^{-1}$, $F(x) = x^4 + 2x^3 - 2x^2 + 2x + 1$, which we can compare to the $4$ cyclotomic polynomials of degree $4$ and see that it matches none, so $\cos^{-1}(\varphi^{-1})$ is not a rational multiple of $\pi$.