Say $\mu$ is a Borel probability measure on $\mathbb{R}$ such that for some $t\neq 0$, we have $\hat{\mu}(t)=1$. That means $$1 = \int\limits_{-\infty}^{\infty} e^{-itx}d\mu(x). $$
Does it then follow that $\mu$ must be concentrated on $\left\lbrace \frac{2\pi n}{t} : n \in \mathbb{Z} \right\rbrace$?
Sorry if you find this question offensive; I failed trigonometry.
Taking real parts we get $1=\int \cos (tx) d\mu (x)$ which can be written as $\int [1-\cos (tx)] d\mu(x)=0$. Since the integrand is non-negative this implies $1-\cos (tx)=0$ for almost all $x$. Can you complete the argument ?