\({\log _3}\left( {\dfrac{1}{3}-\left| {\dfrac{{3\pi }}{2}-x} \right|} \right) = \sin x\)
Так как \(\dfrac{1}{3}-\left| {\dfrac{{3\pi }}{2}-x} \right| \le \dfrac{1}{3}\), то левая часть \({\log _3}\left( {\dfrac{1}{3}-\left| {\dfrac{{3\pi }}{2}-x} \right|} \right) \le -1\), а правая часть \(-1 \le \sin x \le 1\). Поэтому равенство возможно, если:
\(\left\{ {\begin{array}{*{20}{c}}{{{\log }_3}\left( {\dfrac{1}{3}-\left| {\dfrac{{3\pi }}{2}-x} \right|} \right) = -1,}\\{\sin x = -1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}x = \dfrac{{3{\rm{\pi }}}}{2},\\x = \dfrac{{3{\rm{\pi }}}}{2} + 2{\rm{\pi }}k,\,\,\,k \in Z\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x = \dfrac{{3{\rm{\pi }}}}{2}.\)
Ответ: \(\dfrac{{3{\rm{\pi }}}}{2}.\)