Let the period of $\cos(\sin(\sin(\frac x3)))$ be $P$, where $P>0$ is the smallest value satisfying the equation $$\cos(\sin(\sin(\frac x3)))=\cos(\sin(\sin(\frac {x+P}{3})))$$ for all values of $x$.
$$\implies \sin(\sin(\frac x3))=\pm\sin(\sin(\frac {x+P}{3}))+k\cdot2\pi, k\in\mathbb{Z}$$
Due to the range of $\sin$ this equation has no solution unless $k=0$. So now we solve
$$\sin(\sin(\frac x3))=\pm\sin(\sin(\frac {x+P}{3}))=\sin(\pm\sin(\frac {x+P}{3}))$$
So either $$\sin(\frac x3)=\pm\sin(\frac {x+P}{3})+k'\cdot
2\pi,k'\in\mathbb{Z}$$
or
$$\sin(\frac x3)=\pi\mp\sin(\frac {x+P}{3})+k'\cdot
2\pi,k'\in\mathbb{Z}$$
Again, due to the range of $\sin$, we must have $k'=0$ in each case, and for the same reason we can ignore the second pair of cases.
So we are reduced to solving $$\sin(\frac x3)=\pm\sin(\frac {x+P}{3})=\sin(\pm\frac {x+P}{3})$$
$$\implies \frac x3=\pm\frac{x+P}{3}+k''\cdot2\pi,k''\in\mathbb{Z}$$
or
$$\frac x3=\pi\mp\frac{x+P}{3}+k''\cdot2\pi,k''\in\mathbb{Z}$$
Of these four options, only two can be valid for all values of $x$, i.e. the ones where $x$ is eliminated:
The first one with a $+$ sign gives $$-\frac P3=k''2\pi\implies P=6\pi$$
And the second one with a $-$ sign gives $$0=\pi+\frac P3+k''2\pi\implies P=3\pi$$
So the period is $3\pi$.
You might like to try the first one yourself.
Best Answer
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 $T = 2\pi$.