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Our goal is the evaluation of
\begin{equation}\bbox[#ffe,15px,border:1px groove navy]{\ds{%
\sum_{n = 0}^{\infty}\braces{%
{\cos\pars{\bracks{6n + 1}\theta} \over 6n + 1} -
{\cos\pars{\bracks{6n + 5}\theta} \over 6n + 5}}
=
\left.\Re\sum_{n = 0}^{\infty}\pars{{x^{6n + 1} \over 6n + 1} -
{x^{6n + 5} \over 6n + 5}}\right\vert_{\ R\ =\ 1^{-}}}}
\label{1}\tag{1}
\end{equation}
where $\ds{x \equiv R\expo{\ic\theta}}$.
Note that
\begin{align}
&\left.\sum_{n = 0}^{\infty}\pars{%
{x^{6n + 1} \over 6n + 1} - {x^{6n + 5} \over 6n + 5}}
\right\vert_{\ R\ =\ 1_{-}} =
\sum_{n = 0}^{\infty}\bracks{\pars{%
{x^{n + 1} \over n + 1} - {x^{n + 5} \over n + 5}}{1 \over 6}
\sum_{r\ \in\ \Omega}r^{n}}_{\ R\ =\ 1^{-}}\label{2}\tag{2}
\end{align}
where $\ds{\braces{r}}$ are $\ds{\pars{~z^{6} = 1~}}$-roots such that
$\ds{r \in \Omega \equiv
\braces{\exp\pars{k\,{\pi\ic \over 3}}\ {\Large \mid}\ k = 0,1,\ldots,5}}$. For notation simplicity; we switch back and forth, along the result derivation, from $\ds{\sum_{r\ \in\ \Omega}}$ to $\ds{\sum_{k = 0}^{5}}$.
From identity \eqref{2}:
\begin{align}
&\left.\sum_{n = 0}^{\infty}\pars{%
{x^{6n + 1} \over 6n + 1} - {x^{6n + 5} \over 6n + 5}}
\right\vert_{\ R\ =\ 1^{-}} =
\bracks{{1 \over 6}\sum_{r\ \in\ \Omega}{1 \over r}
\sum_{n = 1}^{\infty}{\pars{rx}^{n} \over n} -
{1 \over 6}\sum_{r\ \in\ \Omega}{1 \over r^{5}}
\sum_{n = 5}^{\infty}{\pars{rx}^{n} \over n}}_{\ R\ =\ 1^{-}}
\\[1cm] = &\
\bracks{-\,{1 \over 6}\sum_{r\ \in\ \Omega}r^{5}\ln\pars{1 - xr} +
{1 \over 6}\sum_{r\ \in\ \Omega}r\ln\pars{1 - xr}}_{\ R\ =\ 1^{-}}
\\[5mm] &\phantom{=}+ {1 \over 6}
\underbrace{\sum_{r\ \in\ \Omega}\pars{rx +
{r^{2}x^{2} \over 2} + {r^{3}x^{3} \over 3} + {r^{4}x^{4} \over 4}}}_{\ds{=\ 0}}
=
\left.-\,{1 \over 6}\sum_{r\ \in\ \Omega}\pars{r^{5} - r}\ln\pars{1 - xr}
\,\right\vert_{\ R\ =\ 1^{-}}
\end{align}
Note that, in the limit $\ds{R \to 1^{-}}$, there isn't any contribution from
$\ds{r = 1}$ to the above sum such that we can set $\ds{R = 1}$ in the RHS of the following expressions:
\begin{align}
&\left.\sum_{n = 0}^{\infty}\pars{%
{x^{6n + 1} \over 6n + 1} - {x^{6n + 5} \over 6n + 5}}
\right\vert_{\ R\ =\ 1^{-}} =
-\,{1 \over 6}\sum_{r\ \in\ \Omega}r^{3}\pars{r^{2} - r^{-2}}\ln\pars{1 - xr}
\\[5mm] = &\
-\,{1 \over 6}\sum_{k = 0}^{5}\exp\pars{3\,{k\pi \over 3}\ic}
\bracks{2\ic\,\Im\pars{\exp\pars{2\,{k\pi \over 3}\ic}}}
\ln\pars{1 - \exp\pars{\bracks{\theta + {k\pi \over 3}}\ic}}
\\[5mm] = &\
-\,{1 \over 3}\,\ic\sum_{k = 0}^{5}\pars{-1}^{k}\sin\pars{2k\pi \over 3}
\ln\pars{1 - \exp\pars{\ic\phi_{k}}}\qquad\mbox{where}\qquad
\phi_{k} \equiv \theta + {1 \over 3}\,k\pi\label{3}\tag{3}
\end{align}
\begin{align}
&\mbox{Then,}\quad\left.\sum_{n = 0}^{\infty}\pars{%
{x^{6n + 1} \over 6n + 1} - {x^{6n + 5} \over 6n + 5}}
\right\vert_{\ R\ =\ 1^{-}}
\\[5mm] = &\
-\,{1 \over 3}\,\ic\sum_{k = 0}^{5}\pars{-1}^{k}\sin\pars{2k\pi \over 3}
\ln\pars{\vphantom{\Huge A}\exp\pars{{1 \over 2}\,\phi_{k}\ic}\bracks{%
\exp\pars{-\,{1 \over 2}\,\phi_{k}\ic} - \exp\pars{{1 \over 2}\,\phi_{k}\ic}}}
\\[5mm] = &\
-\,{1 \over 3}\,\ic\sum_{k = 0}^{5}\pars{-1}^{k}\sin\pars{2k\pi \over 3}
\ln\pars{\vphantom{\Huge A}\exp\pars{{\phi_{k} + \pi \over 2}\,\ic}
\bracks{2\sin\pars{-\,{\phi_{k} \over 2}}}}
\end{align}
With the last expression and identity \eqref{1} it's clear that
$\pars{~\vphantom{\large A}\mbox{see}\ \phi_{k}\ \mbox{definition in}\ \eqref{3}~}$:
\begin{align}
&\sum_{n = 0}^{\infty}\braces{%
{\cos\pars{\bracks{6n + 1}\theta} \over 6n + 1} -
{\cos\pars{\bracks{6n + 5}\theta} \over 6n + 5}} =
{1 \over 6}\sum_{k = 0}^{5}\pars{-1}^{k}\sin\pars{2k\pi \over 3}
\pars{\phi_{k} + \pi}
\\[5mm] = &\
{1 \over 18}\,\pi\
\underbrace{\sum_{k = 0}^{5}\pars{-1}^{k}\sin\pars{2k\pi \over 3}k}
_{\ds{=\ 3\root{3}}}\ +\
{1 \over 6}\pars{\theta + \pi}\
\underbrace{\sum_{k = 0}^{5}\pars{-1}^{k}\sin\pars{2k\pi \over 3}}
_{\ds{=\ 0}}\ =\
\bbox[10px,#ffe,border:1px groove navy]{\ds{{\root{3} \over 6}\,\pi}}
\end{align}
