How to find period of this periodic function?

Solution 1:

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 $T = 2\pi$.

Solution 2:

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.

Solution 3:

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

Solution 4:

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 $2\pi$.

Solution 5:

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 $2 \pi$.