I am getting a strange result. I have a periodic function $x(t) = 3cos(4t)+2sin(6t)+cos(10t)$. I worked out that the period is $\pi$, but I also worked out that the frequency of the function is 60 hertz. This doesn't make any sense because the frequency should just be $\frac{1}{2\pi}$. What's wrong?
$2\pi/4,2\pi/6,2\pi/10$ are the periods, so the least common multiple is $\pi$. On the other hand, $4\pi,5\pi,10\pi$ are the frequencies so the least common multiple is $60\pi$.
You are right in that the period is $T=\pi$. Thus we have the frequency: $f=\frac{1}{\pi}$ and the angular frequency: $\omega=2\pi f=\frac{2\pi}{T}=2$.
Here is the graph:
also: the functions $cos(4t)$, $sin(6t)$, $cos(10t)$ have $$ \begin{array}{l|ccc} & cos(4t) & sin(6t) & cos(10t) \\ \hline \omega=2\pi f = \frac{2\pi}{T} & 4 & 6 & 10 \\ f=\frac{1}{T} & \frac{2}{\pi} & \frac{3}{\pi} & \frac{5}{\pi} \\ T & \frac{\pi}{2} & \frac{\pi}{3} & \frac{\pi}{5} \\ \end{array} $$ correspondingly.