\(\sin \dfrac{{\pi \left( {2x-3} \right)}}{6} = -0,5\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\dfrac{{\pi \left( {2x-3} \right)}}{6} = -\dfrac{\pi }{6} + 2\pi n\left| { \cdot 6} \right.}\\{\dfrac{{\pi \left( {2x-3} \right)}}{6} = -\dfrac{{5\pi }}{6} + 2\pi n\left| { \cdot 6} \right.}\end{array}} \right.\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\pi \left( {2x-3} \right) = -\pi + 12\pi n\left| {:\pi } \right.}\\{\pi \left( {2x-3} \right) = -5\pi + 12\pi n\left| {:\pi } \right.}\end{array}} \right.\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{2x-3 = -1 + 12n}\\{2x-3 = -5 + 12n}\end{array}} \right.\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{2x = 2 + 12n\left| {:2\,\,} \right.}\\{2x = -2 + 12n\left| {:2} \right.}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 1 + 6n;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x = -1 + 6n,\,\,\,n\, \in \,Z.}\end{array}} \right.\)
Рассмотрим \(x = 1 + 6n,\,\,n\, \in \,Z\). Если \(n = 0\), то \(x = 1\); если \(n = -1\), то \(x = -5\).
Рассмотрим \(x = -1 + 6n,\,\,n\, \in \,Z\). Если \(n = 1\), то \(x = 5\); если \(n = 0\), то \(x = -1\).
Следовательно, наибольший отрицательный корень \(x = -1\).
Ответ: – 1.