Answer
$-5x+\dfrac{5x^3}{3!}-\dfrac{5x^5}{5!}+\dfrac{5x^7}{7!}+...$
Work Step by Step
Consider the Maclaurin Series for $\sin x$ as follows:
$ \sin x=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...$
Plug $- x$ instead of $x$ in the above series and multiply with the term $5$ as follows:
$ 5\sin(- x)=5[(-x)-\dfrac{(-x)^3}{3!}+\dfrac{(-x)^5}{5!}-\dfrac{(-x)^7}{7!}+...]$
$\implies -5x+\dfrac{5x^3}{3!}-\dfrac{5x^5}{5!}+\dfrac{5x^7}{7!}+...$
