How to find fundamental frequency of Fourier Series
Your solution is correct. Obviously, any periodic function with period $T$ is also periodic with period $kT$, $k\in\mathbb{Z}^+$. In the given example, the smallest possible period is $T=1/5$, which is equivalent to $\omega_0=10\pi$. There is no obvious reason why one should choose $T^\prime=kT$ and $\omega_0^{\prime}=\omega_0/k$, with $k&gt;1$. Nevertheless, such a choice, even though unmotivated, is equally correct, because the resulting Fourier series are identical:

$$\begin{align}x(t)&amp;=\sum_nc_ne^{jn\omega_0t}\\&amp;=\sum_nc_n^{\prime}e^{jn\omega_0^{\prime}t}\\&amp;=\sum_nc_n^{\prime}e^{j\frac{n}{k}\omega_0t}\\&amp;=\sum_mc_{km}^{\prime}e^{jm\omega_0t}\end{align}\tag{1}$$

With $\omega_0^{\prime}=\omega_0/k$ we get

$$c^{\prime}_n=\begin{cases}c_{n/k},&amp;n=mk\\0,&amp;\textrm{otherwise}\end{cases}$$