Answer
$\Sigma_{n=0}^\infty \dfrac{(-1)^n 3^{(2n+1)}x^{(2n+1)}}{(2n+1)!}$ or, $ 3x-\dfrac{3^3}{3!} x^3+\dfrac{3^5}{5!}x^5$
Work Step by Step
Consider the Maclaurin Series for $\sin x$ as follows:
$ \sin x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}$
Plug $3x$ instead of $x$
This implies that $ \sin 3x=\Sigma_{n=0}^\infty \dfrac{(-1)^n 3^{(2n+1)}x^{(2n+1)}}{(2n+1)!}$ or, $ 3x-\dfrac{3^3}{3!} x^3+\dfrac{3^5}{5!}x^5$
