\(\arcsin \left( {x\left( {x + y} \right)} \right) + \arcsin \left( {y\left( {x + y} \right)} \right) = \pi \,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\arcsin \left( {x\left( {x + y} \right)} \right) = \pi -\arcsin \left( {y\left( {x + y} \right)} \right).\)
\(\sin \left( {\arcsin \left( {x\left( {x + y} \right)} \right)} \right) = \sin \left( {\pi -\arcsin \left( {y\left( {x + y} \right)} \right)} \right);\)
\(x\left( {x + y} \right) = \sin \left( {\arcsin \left( {y\left( {x + y} \right)} \right)} \right);\,\,\,\,\,\,\,\,x\left( {x + y} \right) = y\left( {x + y} \right);\)
\(x\left( {x + y} \right) = y\left( {x + y} \right)\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left( {x + y} \right)\left( {x-y} \right) = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x + y = 0,\\x-y = 0.\end{array} \right.\)
Если \(x + y = 0\), то \(\arcsin 0 + \arcsin 0 \ne \pi \), то есть равенство не выполняется.
Если \(x-y = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,y = x\), то:
\(\arcsin 2{x^2} + \arcsin 2{x^2} = \pi \,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\arcsin 2{x^2} = \dfrac{\pi }{2}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,2{x^2} = 1\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = \dfrac{{\sqrt 2 }}{2},\\x = -\dfrac{{\sqrt 2 }}{2}.\end{array} \right.\)
Так как \(y = x,\) то при \(x = \dfrac{{\sqrt 2 }}{2}\) получим \(y = \dfrac{{\sqrt 2 }}{2},\) а при \(x = -\dfrac{{\sqrt 2 }}{2}\) получим \(y = -\dfrac{{\sqrt 2 }}{2}.\)
Ответ: \(\left( {\dfrac{{\sqrt 2 }}{2};\dfrac{{\sqrt 2 }}{2}} \right),\,\,\,\,\,\left( {-\dfrac{{\sqrt 2 }}{2};-\dfrac{{\sqrt 2 }}{2}} \right).\)