Answer
The given series converges for all of $x$ and the sum is $\frac{3}{3-sinx}$.
Work Step by Step
$\Sigma^{\infty}_{n=0} \frac{sin^{n}x}{3^{n}} = \Sigma^{\infty}_{n=0} (\frac{sinx}{3})^{n}$
$a=1$ and $r=\frac{sinx}{3}$
The series converges when
$|r| \lt 1$
$|\frac{sinx}{3}| \lt 1$
$-3 \lt sinx \lt 3$
Since $|sinx| \leq 1$ for all x, we should have $-3 \lt sinx \lt 3$ for all $x$ and so the given series converges for all $x$.
$\Sigma^{\infty}_{n=0} \frac{sin^{n}x}{3^{n}} = \frac{a}{1-r}$
$=\frac{1}{1-\frac{sinx}{3}}$
$=\frac{3}{3-sinx}$
