\(\arcsin \left( {3{x^2}-4x-1} \right) = -\arcsin \left( {-x-1} \right)\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\arcsin \left( {3{x^2}-4x-1} \right) = \arcsin \left( {x + 1} \right)\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{-1 \le x + 1 \le 1,\,\,\,\,\,\,\,\,\,\,\,\,}\\{3{x^2}-4x-1 = x + 1}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{-2 \le x \le 0,\,\,\,\,\,\,\,\,\,\,\,\,}\\{3{x^2}-5x-2 = 0}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{-2 \le x \le 0,}\\{\left[ {\begin{array}{*{20}{c}}{x = 2,}\\{x = -\dfrac{1}{3}}\end{array}\,\,\,\,\,} \right.}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x = -\dfrac{1}{3}.\)
Ответ: \(-\dfrac{1}{3}.\)