\(\left( {{x^2}-x-6} \right)\left( {-{x^2} + 1} \right) \ge 0.\)
Решим неравенство методом интервалов:
\(\left( {{x^2}-x-6} \right)\left( {-{x^2} + 1} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{x^2}-x-6 = 0,}\\{-{x^2} + 1 = 0\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 3,\,\,\,\,}\\{x = -2,}\\{x = 1,\,\,\,\,}\\{x = -1.}\end{array}} \right.\)
Следовательно, решение исходного неравенства: \(x\, \in \,\left[ {-2;-1} \right] \cup \left[ {1;3} \right].\)
Ответ: \(\left[ {-2;\;-1} \right] \cup \left[ {1;\;3} \right].\)