The points at which $(2\pi n+x)\sin(2\pi n+x)=1$ are the the points where
$ \begin{align} 0 &=\frac1{2\pi n+x}-\sin(x)\\ &=\left(\frac1{2\pi n}-\frac x{4\pi^2 n^2}+\frac{x^2}{8\pi^3n^3}-\frac{x^3}{16\pi^4n^4}+\dots\right)-\left(x-\frac{x^3}6+\dots\right)\\ &=\frac1{2\pi n}-\frac{4\pi^2n^2+1}{4\pi^2n^2}x+\frac1{8\pi^3n^3}x^2+\frac{8\pi^4n^4-3}{48\pi^4n^4}x^3+O\left(x^4\right)\tag{1} \end{align} $ Thus, $ \frac{2\pi n}{4\pi^2n^2+1} =x-\frac1{8\pi^3n^3+2\pi n}x^2-\frac{8\pi^4n^4-3}{48\pi^4n^4+12\pi^2n^2}x^3+O\left(x^4\right)\tag{2} $ The inverse series for the series on the right-hand-side of $(2)$ is $ x=y+\frac1{8\pi^3n^3+2\pi n}y^2+\frac{32\pi^6n^6+8\pi^4n^4-12\pi^2n^2+3}{12(4\pi^3n^3+\pi n)^2}y^3+O\left(y^4\right)\tag{3} $ Plugging $y=\frac1{2\pi n}-\frac1{8\pi^3n^3}+O\left(\frac1{n^5}\right)$ from the left-hand-side of $(2)$ into $(3)$ yields $ x=\frac1{2\pi n}-\frac5{48\pi^3n^3}+O\left(\frac1{n^4}\right)\tag{4} $ The root in question is located at $ 2\pi n+x=2\pi n+\frac1{2\pi n}-\frac5{48\pi^3n^3}+O\left(\frac1{n^4}\right)\tag{5} $ and $(5)$ says that $L=\frac1{2\pi}$ and $L_2=-\frac5{48\pi^3}$ .
A second approach (similar to Ewan Delanoy)
There are two sequences of roots; one near $2n\pi$ $ \begin{align} 1&=(2n\pi+x)\sin(2n\pi+x)\\ \frac1{2n\pi}&=\left(1+\frac{x}{2n\pi}\right)\sin(x)\tag{6} \end{align} $ and one near $(2n+1)\pi$ $ \begin{align} 1&=((2n+1)\pi+x)\sin((2n+1)\pi+x)\\ -\frac1{(2n+1)\pi}&=\left(1+\frac{x}{(2n+1)\pi}\right)\sin(x)\tag{7} \end{align} $ To solve either $(6)$ or $(7)$, we will use the series $ (1+px)\sin(x)=x+px^2-\frac16x^3-\frac p6x^4+O\left(x^5\right)\tag{8} $ and its inverse $ x=y-py^2+\frac{1\color{#C00000}{+12p^2}}{6}y^3\color{#C00000}{-\frac{2p+15p^3}{3}y^4}+O\left(y^5\right)\tag{9} $ For $(6)$, we will use $y=p=\frac1{2n\pi}$, and for $(7)$, we will use $y=-p=-\frac1{(2n+1)\pi}$. For each of these, the terms in red contribute no more than the error term, so can be ignored.
Thus, for the $x$ in $(6)$, we get $ x=p-\frac56p^3+O\left(p^5\right)\tag{10} $ and for the $x$ in $(7)$, we get $ x=-p-\frac76p^3+O\left(p^5\right)\tag{11} $ Thus, the roots for $(6)$ are $ 2n\pi+x=2n\pi+\frac1{2n\pi}-\frac5{48n^3\pi^3}+O\left(\frac1{n^5}\right)\tag{12} $ and the roots for $(7)$ are $ (2n+1)\pi+x=(2n+1)\pi-\frac1{(2n+1)\pi}-\frac7{6(2n+1)^3\pi^3}+O\left(\frac1{n^5}\right)\tag{13} $ The roots in $(12)$ are the same as those in $(5)$ only with better error term. We can continue from there.