For a CT signal
x(t) $ = \sum_{k=-\infty}^\infty a_k e^{jk\omega t} $
Where
x(t) $ = 3 + 2cos(4\pi t) = 3 + (e^{j4\pi t} + e^{-j4\pi t} ) $
$ \omega = 4\pi $
A signal $ e^{jk\omega t} $ is periodic if and only if $ \left (\frac{\omega}{2\pi} \right) $ is a rational number
$ \left ( \frac{4\pi}{2\pi} \right ) = \left ( \frac{2}{1} \right ) $
2 is a rational number!
$ a_o = 3 $
$ a_4 = 1 $
$ a_{-4} = 1 $
