This looks ugly. To simplify it a bit, notice that the LHS is equal to ##\frac12 \sin (2x)##. In particular, this lies in ##\left[-\frac12,\frac12\right]##, so that any ##x## solving your equation would have to belong to ##\left[\frac1{\sqrt{e}},\sqrt{e}\right]##.
Using some knowledge of the functions ##\sin## and ##\ln## (i.e. knowing where each of them is positive/negative and where each of them is positively/negatively sloped, and using that ##2\in (0,\pi)##), you can use intermediate value theorem to show that the equation has a unique solution ##x^*>0##, and that it satifies ##x^*\in (1, \sqrt{e})##.
As for an explicit solution, good luck.