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?
 A: 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.
A: To find a period is no problem, note that $2\pi$ is a period of both parts. 
A: 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 $T = 2\pi$.
A: The (smallest) period can be shown to be $2\pi$ by yet another method. Note that
$$f(x)=2\sin(3x)+3\sin(2x)=2\sin x(\cos x+1)(4 \cos x-1).$$
So the zeros of $f(x)$ in $[0,2\pi]$ occur at $0, \pi, 2\pi$ and at two other points obtained from $4 \cos x-1=0$ which are $\cos^{-1}(1/4)$ and $2\pi-\cos^{-1}(1/4).$
In any period of $f(x)$ this pattern of zeros would have to repeat. However the numeric estimates $$0.00,\ 1.318,\ 3.141,\ 4.965,\ 6,28$$ are not evenly spaced, in fact the spacings are
$$1.318,\ 1.823,\ 1.823,\ 1.318.$$
It seems clear this uneven spacing of the zeros makes it impossible for $f(x)$ to repeat before a full $2\pi$ goes by.
EDIT: It is because this spacing pattern is of the form ABBA(where lengths A,B different) that the period must be $2\pi$ and not smaller. A different function with zero spacing pattern ABAB might have period $\pi$. If extended, the ABBA pattern is ABBAABBAABBA.., and there are no periods of less than four letters length in this pattern. The repeated pattern ABABABAB.. has a period of two letters, hence zero spacing of that form might have period $\pi$ rather than $2\pi$.
A: The fundamental period of sin function is $2\pi$.
Now,Add the period of $2\pi$ in each sin function of given equation.
$y=2\sin(3x) + 3\sin(2x)$
$y=2\sin(3x+2\pi) + 3\sin(2x+2\pi)$
taking $3$ common from 1st sin fun. and $2$ from 2nd sin fun.
$y=2\sin[3(x+2\pi/3)] + 3\sin[2(x+2\pi/2)] $
so, the period of 1st sin fun. is  $\frac{2\pi}{3}$
and the peiod of 2nd sin fun. is ${\pi}{}$
Finally, find the lowest common multiple of these two periods
which is = $2{\pi}$ Answer.
A: The period of the first is $2 \pi/3$, that of the second is $\pi$. When we take the  sum, the resulting period is the least common multiple of the two, which is $2 \pi$.   
A: the correct ans will be $2\pi$, as the period of first is $2\pi$ by $3$ and second is $2\pi$ by $4$.
A: given f(x) = 2sin3x+3sin2x
we know sinx fundamental period (fp) is 2pi.
Also, fp of sin(ax+b)=2pi/modulus a.
Therefore fp of sin3x=2pi/3 & fp of sin2x=2pi/2=pi
cosider LCM of {2/3 ,1}     (the LCM of fp of sin3x&sin2x)
         =LCM{2,1}/HCF{3,1}=2/3
hence given f(x) is periodic with fp=2pi/3
