How to find period of this periodic function? How can I find a period of this function?

$$2\sin{3x} + 3\sin{2x}$$

Is here any way how to sum both sinuses?
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 $T = 2\pi$.
the correct ans will be $2\pi$, as the period of first is $2\pi$ by $3$ and second is $2\pi$ by $4$.