Finding the value of trignometric series $\Sigma^{\infty}_{0}\dfrac{\cos nx}{3^n}$

Question

$Q. $ It is being given that $\cos x\in(0,\frac{\pi}{2})$ and $\cos(x) = \frac{1}{3}$ we need to find $$\Sigma^{\infty}_{0}\dfrac{\cos nx}{3^n}$$

I tried writing some terms, and then multiplying by three and subtracting original from it, but that took me nowhere!

Then I tried using expansions of $\cos(nx)$ but realised this wasn't needed here and made it more unsolvable.

I also wanted to know if there was a more general method to geometric-trigonometric series like these

Answer 1

Hint. One may write, for $x \in \mathbb{R}$, $$ \dfrac{\cos nx}{3^n}=\text{Re}\left(\dfrac{e^{ix}}{3}\right)^n= \text{Re} \: z^n $$ with $\displaystyle z= \dfrac{e^{ix}}{3}$ then one may recall that $$ \sum_{n=0}^\infty z^n= \frac1{1-z},\quad |z|<1. $$