The original problem is from the Problem Set 7 of MIT OpenCourseware: Find the Fourier series coefficients for $$ x(t)=sin(10\pi t+\frac{\pi}{6}) $$ What I did is to rewrite it in exponential form $\frac{1}{2j}e^{j\frac{\pi}{6}}e^{j10\pi t}-\frac{1}{2j}e^{-j\frac{\pi}{6}}e^{-j10\pi t}$, and take $\omega_0=10\pi$ as the fundamental frequency. The non-zero coefficients I got are $a_1=\frac{1}{2j}e^{j\frac{\pi}{6}}$ and $a_{-1}=-\frac{1}{2j}e^{-j\frac{\pi}{6}}$.
However, the solution takes $\omega_0=2\pi$, which gives $a_5=\frac{1}{2j}e^{j\frac{\pi}{6}}$ and $a_{-5}=-\frac{1}{2j}e^{-j\frac{\pi}{6}}$.
I can't find a proper explanation for this problem. What should I take as fundamental frequency? Can the fundamental frequency be arbitary number for continuous time Fourier series?
Here is the link of the problem set: Problem set 7
The link of the solution: Solution