Задача 10. Решите уравнение: \(\arcsin \left( {x\sqrt 2 } \right) + \arccos x = \dfrac{\pi }{4}.\)
ОТВЕТ: \(-\dfrac{{\sqrt 2 }}{2}.\)
\(\arcsin \left( {x\sqrt 2 } \right) + \arccos x = \dfrac{\pi }{4}.\) Запишем ОДЗ: \(\left\{ {\begin{array}{*{20}{c}}{-1 \le x\sqrt 2 \le 1,}\\{-1 \le x \le 1\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,-\dfrac{{\sqrt 2 }}{2} \le } \right.x \le \dfrac{{\sqrt 2 }}{2}.\) \(\sin \left( {\arcsin \left( {x\sqrt 2 } \right) + \arccos x} \right) = \sin \dfrac{\pi }{4}.\) Воспользуемся формулой синуса суммы: \(\sin \left( {\alpha + \beta } \right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta .\) \(\sin \left( {\arcsin \left( {x\sqrt 2 } \right) + \arccos x} \right) = \sin \dfrac{\pi }{4}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\sin \left( {\arcsin \left( {x\sqrt 2 } \right)} \right)\cos \left( {\arccos x} \right) + \cos \left( {\arcsin \left( {x\sqrt 2 } \right)} \right)\sin \left( {\arccos x} \right) = \dfrac{{\sqrt 2 }}{2}\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,x\sqrt 2 \cdot x + \sqrt {1-2{x^2}} \sqrt {1-{x^2}} = \dfrac{{\sqrt 2 }}{2}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,2\sqrt {1-2{x^2}} \sqrt {1-{x^2}} = \sqrt 2 \left( {1-2{x^2}} \right)\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,2\sqrt {1-2{x^2}} \sqrt {1-{x^2}} -\sqrt 2 {\left( {\sqrt {1-2{x^2}} } \right)^2} = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\sqrt {1-2{x^2}} \left( {2\sqrt {1-{x^2}} -\sqrt 2 \sqrt {1-2{x^2}} } \right) = 0\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sqrt {1-2{x^2}} = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{2\sqrt {1-{x^2}} = \sqrt 2 \cdot \sqrt {1-2{x^2}} }\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \pm \dfrac{{\sqrt 2 }}{2},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{2-2{x^2} = 1-2{x^2}}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x = \pm \dfrac{{\sqrt 2 }}{2}.\) Проверкой убеждаемся, что \(x = -\dfrac{{\sqrt 2 }}{2}\) подходит, а \(x = \dfrac{{\sqrt 2 }}{2}\) не подходит. Ответ: \(-\dfrac{{\sqrt 2 }}{2}\).