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