How to find period of this periodic function?

You can always go the hard way analyzing, for arbitrary $T$: $$ 2 \sin(3 (x+T) ) + 3 \sin(2(x+T)) = \\ 2 \sin(3x) \cos(3T) + 2 \cos(3x) \sin(3T) + 3 \sin(2x) \cos(2T) + 2 \cos(2x) \sin(2T) $$ When $T$ equals a period you should have $$ \sin(3T) = 0, \quad \sin(2T) = 0, \quad \cos(3T) = 1, \quad \cos(2T) = 1 $$ Since $\sin(3T) = \sin(T+2T) = \sin(T) \underbrace{\cos(2T)}_1 + \cos(T) \underbrace{\sin(2T)}_0 = \sin(T)$, and $\cos(3T) = \cos(T) \cos(2T) - \sin(T) \sin(2T) = \cos(T)$. We have: $$ \sin(T) = 0, \quad \sin(2T) = 0, \quad \cos(T) = 1, \quad \cos(2T) = 1 $$ Repeating the exercise we see that $\sin(2T) = 0, cos(2T)=1$ is implied by $\sin(T)=0$ and $\cos(T) = 1$. Solving $\sin(T)=0$ and $\cos(T)=1$ is easy. There are infinitely many solutions: $$ T = 2 \pi n, \quad n \in \mathbb{Z} $$ The minimal solution is $T = 2\pi$.

One has period $\pi$ and the other has period $2\pi /3$. What you want now is to see when they "match up". This is obtained in $2\pi$. Basically, this is $3\times 2\pi/3$ and $2\times \pi$. We're just cross multiplying periods.

To find a period is no problem, note that $2\pi$ is a period of both parts.