Is $t\mapsto \left|\cos (t)\right|$ a characteristic function?

Factoid 1: If a characteristic function is infinitely differentiable at zero, all the moments of the corresponding random variable are finite.

Factoid 2: If all the moments of a random variable are finite, the corresponding characteristic function is infinitely differentiable everywhere on the real line.

Factoid 3: The function $t\mapsto|\cos(t)|$ is infinitely differentiable at $t=0$ but not everywhere on the real line, for example not at $t=\pi/2$.

Ergo.

Since $\phi(t) = | \cos(t) |$ is periodic with period $\pi$ and even and if it is valid, it should correspond to a symmetric discrete random variable.

It is not hard to establish that: $$ | \cos(t) | = \frac{2}{\pi} + \frac{4}{\pi} \sum_{m=1}^\infty \frac{(-1)^{m-1}}{4 m^2-1} \cos(2 m t) $$

Comparing this to $\phi_X(t) = \sum_{m=-\infty}^\infty c_m \mathrm{e}^{i t m}$ we see that $c_4 = - \frac{2}{\pi} \cdot \frac{1}{15}$ is negative, thus can not be a probability of any random variable.

W. Feller, An Introduction to Probability Theory and Applications, Volume I, XIX.4, Theorem 1.
A continuous function $\phi$ with period $2\pi$ is a characteristic function iff its Fourier coefficients (4.2) satisfy $\phi_k \ge 0$ and $\phi(0) = 1$.

$$ \phi_k = \frac{1}{2\pi}\int_{-\pi}^{\pi} e^{-i k \zeta} \phi(\zeta)\,d\zeta \tag{4.2} $$