Fourier series of $\log(a +b\cos(x))$? By numerical computation it seems like, if $a_0 < a_1$:
$$
\begin{multline}
\log({a_0}^2 + {a_1}^2 + 2 a_0 a_1 \cos(\omega t)) = \log({a_0}^2 + {a_1}^2) \\
+ \frac{a_0}{a_1}\cos(\omega t) 
- \frac{1}{2}\frac{{a_0}^2}{{a_1}^2}\cos(2\omega t) 
+ \frac{1}{3}\frac{{a_0}^3}{{a_1}^3}\cos(3\omega t) 
- \frac{1}{4}\frac{{a_0}^4}{{a_1}^4}\cos(4\omega t) \ldots 
\end{multline}
$$
If $a_0 > a_1$, the two term must be exchanged. 
I'm quite confident in this solution but I cannot find a way to prove it mathematically...
Would be nice to know the reason of this result!
 A: To apply the solution proposed by @ChristianRemling to the more general case stated in the title you need to do the following:
The formula implies $a > 0$ and $|b| < a$. To solve the problem, we reformulate the formula as:
$$
\begin{align}
r|1 + qe^{ix}|^2
&= r(1 + qe^{ix})(1 + qe^{-ix}) \\
&= r(1 + 2q\cos x + q^2) \\
& = a + b\cos x \\
\end{align}
$$
This requires to solve the following system:
$$
\begin{align}
a &= r(1 + q^2) \\
b &= 2rq
\end{align}
$$
From the two possible solutions we take the one where $|q|<1$ which is mandatory for the Taylor expansion. 
$$
\begin{align}
q = \frac{a - \sqrt{a^{2} - b^{2}}}{b} \\
r = \frac{a + \sqrt{a^{2} - b^{2}}}{2}
\end{align}
$$
Finally, Taylor series expansion solve the problem:
$$
\begin{align}
\log (a + b\cos x)
&= \log r 
+ 2\log | 1 + qe^{ix} | \\
&= \log r 
+ 2 \sum {(-1)}^{n-1} \frac{q^n}{n} e^{inx} \\
&= \log r 
+ 2 \sum {(-1)}^{n-1} \frac{q^n}{n} \cos nx
\end{align}
$$
A: Let's consider
$$
f(x) = \log (1+q^2+2q\cos x) = \log |1+qe^{ix}|^2 ,
$$
which differs from your function only by the additive constant $2\log a_1$ if we take $q=a_0/a_1$. Since $|q|<1$, we can use the Taylor series of $\log(1+z)$ to write
$$
\begin{align}
f(x)  = 2\,\textrm{Re}\; \log (1+qe^{ix}) = 2\,\textrm{Re}\sum_{n\ge 1} (-1)^{n-1}\frac{q^n}{n} e^{inx} \\
= 2\sum_{n\ge 1} (-1)^{n-1}\frac{q^n}{n} \cos nx .
\end{align}
$$
