CT signal
$ x(t) = 2 + cos({\frac{2\pi t}{3}})+ 4sin({\frac{5\pi t}{3}})\, $
Coefficients
$ cos({\frac{2\pi t}{3}}) = \frac{1}{2}e^{\frac{j2\pi t}{3}} + \frac{1}{2}e^{\frac{-j2\pi t}{3}} $
$ 4sin({\frac{5\pi t}{3}}) = -2je^{\frac{j5\pi t}{3}} + 2je^{\frac{-j5\pi t}{3}} $
$ x(t) = 2 + \frac{1}{2}e^{\frac{j2\pi t}{3}} + \frac{1}{2}e^{\frac{-j2\pi t}{3}} -2je^{\frac{j5\pi t}{3}} + 2je^{\frac{-j5\pi t}{3}} $
$ x(t) = 2 + \frac{1}{2}e^{\frac{2j2\pi t}{6}} + \frac{1}{2}e^{\frac{-2j2\pi t}{6}} -2je^{\frac{2j5\pi t}{6}} + 2je^{\frac{-2j5\pi t}{6}} $
Then we can know the fundamental frequency is $ \frac{\pi}{3} $.
Also, we can get coefficients $ a_0, a_2, a_5, a_-2, a_-5 $.
$ a_0 = 2, a_2 = a_-2 = \frac{1}{2}, a_5 = -2j, a_-5 = 2j $
