Nice question

Hint

Since $cos(na)$ is a polynomial in $cos(a)$ with integer coefficients, it follows that if $cos(a)$ is rational then so is $cos(na)$ for all integers $n$. From the hypothesis, we deduce that for each positive integers $p,q$ the numbers $cos(p(k-1)x)$ and $cos(qkx)$ are rational. But $k$ and $k-1$ are coprimes, thus there are infinite number of couples of positive integers such that $p(k-1)-qk = 1$. (1) In partcular, we may find $p,q$ such that $p(k-1),qk >k$ satisfying (1). Let $n = p(k-1)$ then $n-1 = qk$ and, from above, the numbers $cos((n-1)x)$ and $cos(nx)$ are rational. Pierre.

For an explicit example, set $n = k^{2}$. Then $n-1$ is a multiple of $k-1$, and $n$ is a multiple of $k$.