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