CT signal
$ x(t) = 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) = \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) = \frac{1}{2}e^{\frac{2j2\pi t}{6}} + \frac{1}{2}e^{\frac{-2j2\pi t}{6}} -2je^{\frac{5j2\pi t}{6}} + 2je^{\frac{-5j2\pi t}{6}} $
Then we can know the fundamental frequency is $ \frac{\pi}{3} $.
Also, we can get coefficients $ a_2 $,$ a_{-2} $,$ a_5 $,
$ a_{-5} $.
$ a_2 = a_{-2} = \frac{1}{2}, a_5 = -2j, a_{-5} = 2j, a_k = 0, $where k is not 2,-2,5,-5
