\(\sin \dfrac{{\pi \left( {2x-1} \right)}}{4} = \dfrac{{\sqrt 2 }}{2}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\dfrac{{\pi \left( {2x-1} \right)}}{4} = \dfrac{\pi }{4} + 2\pi k\left| { \cdot 4} \right.\,\,\,\,}\\{\dfrac{{\pi \left( {2x-1} \right)}}{4} = \dfrac{{3\pi }}{4} + 2\pi k\left| { \cdot 4} \right.}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\pi \left( {2x-1} \right) = \pi + 8\pi k\left| {:\pi } \right.}\\{\pi \left( {2x-1} \right) = 3\pi + 8\pi k\left| {:\pi } \right.}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{2x-1 = 1 + 8k,}\\{2x-1 = 3 + 8k\,}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{2x = 2 + 8k\left| {:2} \right.}\\{2x = 4 + 8k\left| {:2} \right.\,}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 1 + 4k,}\\{x = 2 + 4k,}\end{array}} \right.\,\,\,\,\,\,k \in Z.\)
Рассмотрим \(x = 1 + 4k,\,\,\,k\, \in \,Z\). Если \(k = 0\), то \(x = 1\); если \(k = -1\), то \(x = -3.\)
Рассмотрим \(x = 2 + 4k,\,\,\,k\, \in \,Z\). Если \(k = 0\), то \(x = 2\); если \(k = -1\), то \(x = -2.\)
Следовательно, наибольший отрицательный корень \(x = -2.\)
Ответ: \(-2.\)