Consider the function
$$f(x)=\sum_{n=1}^\infty\frac{x^{3n-1}}{(3n-1)!}$$
Consider this a Maclaurin series for $f(x)$. So we have $f(0)=0,f'(0)=0$ and $f''(0)=1$. Finally, take note that $f'''(x)=f(x)$. Solve this initial value problem and multiply by $x$ to get your answer.
It looks like it may take some more work to get this into the desired form, however, so I'll continue. The characteristic equation for our problem is $s^3-1=0$, whose roots are the third roots of unity $1$ and $-\frac12\pm\frac{\sqrt3}2i$. This suggests
$$f(x)=k_1e^x+k_2e^{-\frac x2}\sin(\frac{x\sqrt3}2)+k_3e^{-\frac x2}\cos(\frac{x\sqrt3}2)=$$
$$k_1e^x+e^{-\frac x2}[k_2\sin(\frac{x\sqrt3}2)+k_3\cos(\frac{x\sqrt3}2)]$$
$$f'(x)=k_1e^x+e^{-\frac x2}[(-\frac12k_2-\frac{\sqrt3}2k_3)\sin(\frac{x\sqrt3}2)+(-\frac12k_3+\frac{\sqrt3}2k_2)\cos(\frac{x\sqrt3}2)]$$
$$f''(x)=k_1e^x+e^{-\frac x2}[(\frac14k_2+\frac{\sqrt3}4k_3+\frac{\sqrt3}4k_3-\frac34k_2)\sin(\frac{x\sqrt3}2)+(\frac14k_3-\frac{\sqrt3}4k_2-\frac{\sqrt3}4k_2-\frac34k_3)\cos(\frac{x\sqrt3}2)]=$$
$$k_1e^x+e^{-\frac x2}[(-\frac12k_2+\frac{\sqrt3}2k_3)\sin(\frac{x\sqrt3}2)+(-\frac12k_3-\frac{\sqrt3}2k_2)\cos(\frac{x\sqrt3}2)]$$
Plugging in initial conditions
$$k_1+k_3=0$$
$$k_1+\frac{\sqrt3}2k_2-\frac12k_3=0$$
$$k_1-\frac{\sqrt3}2k_2-\frac12k_3=1$$
From the first equation, we get $k_3=-k_1$. Adding the second and third gives $k_3=2k_1-1$. So $k_1=\frac13,k_3=-\frac13$ and
$$\frac13+\frac{\sqrt3}2k_2+\frac16=0$$
$$\frac{\sqrt3}2k_2=-\frac12$$
$$k_2=-\frac{\sqrt3}3$$
So the last step is to prove
$$\sqrt3\sin(\frac{x\sqrt3}2)+\cos(\frac{x\sqrt3}2)=2\sin(\frac{x\sqrt3}2+\frac\pi6)$$
This last step can be done by solving $a\cos b=\sqrt3$ and $a\sin b=1$. This gives $\tan b=\frac{\sqrt3}3$ and $a^2=3+1=4$.