= Solution
Use the real <Fourier series> convention $f=a_0/2+\sum_{n\geq1}(a_n\cos nx+b_n\sin nx)$. The <function> is an <even function>, so $b_n=0$. For noninteger real $\mu$,
$$
a_0=\frac2\pi\int_0^\pi\cos(\mu x)\,dx=\frac{2\sin\pi\mu}{\pi\mu}.
$$
For $n\geq1$, the product-to-sum identity gives
$$
\begin{aligned}
a_n&=\frac1\pi\left(\frac{\sin\pi(\mu-n)}{\mu-n}
+\frac{\sin\pi(\mu+n)}{\mu+n}\right)\\
&=\frac{2\mu(-1)^n\sin\pi\mu}{\pi(\mu^2-n^2)}.
\end{aligned}
$$
Thus the <Fourier series of a noninteger-frequency cosine> is
$$
\boxed{f(x)=\frac{\sin\pi\mu}{\pi\mu}
+\frac{2\mu\sin\pi\mu}{\pi}\sum_{n=1}^{\infty}
\frac{(-1)^n\cos nx}{\mu^2-n^2}}.
$$
Its <Fourier coefficients> are $O(n^{-2})$, giving <absolute convergence> and <uniform convergence>. The periodic <function> is continuous at its joins because the two endpoint cosine values coincide, so the <Dirichlet-Jordan convergence theorem> gives this value at every point.
Putting $\mu=1/3$ gives
$$
\boxed{f(x)=\frac{3\sqrt3}{2\pi}-\frac{3\sqrt3}{\pi}
\sum_{n=1}^{\infty}\frac{(-1)^n\cos nx}{9n^2-1}}.
$$
Evaluate at $x=0$, where $f(0)=1$, and multiply by $\pi/\sqrt3$:
$$
\frac\pi{\sqrt3}=\frac32-3\sum_{n=1}^{\infty}\frac{(-1)^n}{9n^2-1}
=\frac32-\sum_{n=1}^{\infty}\frac{(-1)^n}{3n^2-1/3}.
$$
Comparing with the required form gives $\boxed{A=3,\qquad B=3}$.
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