Задача 6. Решите систему уравнений \(\left\{ {\begin{array}{*{20}{c}} {\sin x + \cos y = 1} \\ {x + y = \dfrac{\pi }{3}\,\,\,\,\,\,\,\,\,\,\,\,\,} \end{array}} \right.\)
Ответ
ОТВЕТ: \(\left( {{{\left( {-1} \right)}^n}\arcsin \sqrt {2-\sqrt 3 } -\dfrac{\pi }{{12}} + \pi \,n,\;\dfrac{{5\pi }}{{12}}-{{\left( {-1} \right)}^n}\arcsin \sqrt {2-\sqrt 3 } -\pi \,n} \right),\,\,\,n \in Z.\)
Решение
Из второго уравнения: \(y = \dfrac{\pi }{3}-x.\) Подставим в первое уравнение:
\(\sin x + \cos \left( {\dfrac{\pi }{3}-x} \right) = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\sin x + \sin \left( {\dfrac{\pi }{6} + x} \right) = 1\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,2\sin \dfrac{{x + \dfrac{\pi }{6} + x}}{2}\cos \dfrac{{x-\dfrac{\pi }{6}-x}}{2} = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,2\sin \left( {x + \dfrac{\pi }{{12}}} \right)\cos \dfrac{\pi }{{12}} = 1\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,2\sin \left( {x + \dfrac{\pi }{{12}}} \right) \cdot \dfrac{{\sqrt {2 + \sqrt 3 } }}{2} = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\sin \left( {x + \dfrac{\pi }{{12}}} \right) = \dfrac{1}{{\sqrt {2 + \sqrt 3 } }}\,\,\,\,\,\,\, \Leftrightarrow \,\)
\( \Leftrightarrow \,\,\,\,\,\,\,\sin \left( {x + \dfrac{\pi }{{12}}} \right) = \sqrt {2-\sqrt 3 } \,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = {\left( {-1} \right)^n}\arcsin \sqrt {2-\sqrt 3 } -\dfrac{\pi }{{12}} + \pi n,\,\,\,n\, \in \,Z.\)
Тогда: \(y = \dfrac{{5\pi }}{{12}}-{\left( {-1} \right)^n}\arcsin \sqrt {2-\sqrt 3 } -\pi n.\)
Ответ: \(\left( {{{\left( {-1} \right)}^n}\arcsin \sqrt {2-\sqrt 3 } -\dfrac{{\rm{\pi }}}{{12}} + {\rm{\pi }}\,n,\;\dfrac{{5{\rm{\pi }}}}{{12}}-{{\left( {-1} \right)}^n}\arcsin \sqrt {2-\sqrt 3 } -{\rm{\pi }}\,n} \right),\,\,\,n \in Z.\)