Suppose we have the $4\times 4$ matrix $V$
$$\begin{bmatrix} 1 & \cos{z_0} & \cos{2z_0} & \cos{3z_0}\\ 1 & \cos{z_1} & \cos{2z_1} & \cos{3z_1}\\ 1 & \cos{z_2} & \cos{2z_2} & \cos{3z_2}\\ 1 & \cos{z_3} & \cos{2z_3} & \cos{3z_3} \end{bmatrix}.$$
Obviously the functions are linearly independent. Suppose moreover that $z_0=-z_3$ and that $z_1=-z_2$. Therefore, since $\cos{x}$ is even, we have that $V(1,2)=V(4,2)$ (1), $V(2,2)=V(3,2)$ and so on for each column.
Is this matrix singular?
(1) $V(1,2)$ is the element in row $1$, column $2$.
The matrix will be singular when the rank is less than three: $$ \rho < 3 $$ This will happen when at least one of the column vectors is a linear combination of the remaining column vectors.
This can happen, for example, when $\cos 2z_{0} = \cos z_0$...
Bad examples include $$ z_{k} = 2(k+1)\pi $$