\(\cos \dfrac{{\pi \left( {x-7} \right)}}{3} = \dfrac{1}{2}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\dfrac{{\pi \left( {x-7} \right)}}{3} = \pm \dfrac{\pi }{3} + 2\pi n\,\left| {\, \cdot 3\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\pi \left( {x-7} \right) = \pm \pi + 6\pi n\,\left| {:\pi \,\,\,\,\, \Leftrightarrow } \right.} \right.\)
\( \Leftrightarrow \,\,\,\,\,x-7 = \pm 1 + 6n\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 8 + 6n;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x = 6 + 6n,\,\,\,\,n\,\, \in \,\,Z.}\end{array}} \right.\)
Рассмотрим \(x = 8 + 6n,\,\,n\, \in \,Z\). Если \(n = -1\), то \(x = 2\); если \(n = -2\), то \(x = -4.\)
Рассмотрим \(x = 6 + 6n,\,\,n\, \in \,Z\). Если \(n = -1\), то \(x = 0\); если \(n = -2\), то \(x = -6.\)
Следовательно, наибольший отрицательный корень \(x = -4.\)
Ответ: – 4.