Задача 3. Решите систему уравнений \(\left\{ {\begin{array}{*{20}{c}} {y-x = \dfrac{1}{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ {\cos \,\,\left( {\,\pi \,x} \right)\,\,\cos \,\left( {\,\pi \,y} \right) = \dfrac{{\sqrt 2 }}{2}} \end{array}} \right.\)
Ответ
ОТВЕТ: \(\left( {n,\;\dfrac{1}{4} + n} \right),\,\,\,\left( {-\dfrac{1}{4} + n,\;n} \right),\,\,\,n \in Z.\)
Решение
Из первого уравнения: \(y = x + \dfrac{1}{4}.\) Подставим во второе:
\(\cos \pi x \cdot \cos \left( {\pi \left( {x + \dfrac{1}{4}} \right)} \right) = \dfrac{{\sqrt 2 }}{2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\cos \pi x \cdot \cos \left( {\pi x + \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{2}\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\cos \pi x \cdot \left( {\cos \pi x \cdot \cos \dfrac{\pi }{4}-\sin \pi x \cdot \sin \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{2}\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{\sqrt 2 }}{2}\cos \pi x \cdot \left( {\cos \pi x-\sin \pi x} \right) = \dfrac{{\sqrt 2 }}{2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\cos ^2}\pi x-\cos \pi x\sin \pi x = 1\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,1-{\sin ^2}\pi x-\cos \pi x\sin \pi x = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\sin \pi x \cdot \left( {\sin \pi x + \cos \pi x} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin \pi x = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\sin \pi x + \cos \pi x = 0}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin \pi x = 0,}\\{{\rm{tg}}\pi x = -1}\end{array}} \right.\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\pi x = \pi n,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\pi x = -\dfrac{\pi }{4} + \pi n}\end{array}} \right.\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = n,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x = -\dfrac{1}{4} + n,}\end{array}} \right.\,\,\,n\, \in \,Z.\)
Если \(x = n\), то \(y = \dfrac{1}{4} + n\); если \(x = -\dfrac{1}{4} + n\), то \(y = n.\)
Ответ: \(\left( {n,\;\dfrac{1}{4} + n} \right),\,\,\,\left( {-\dfrac{1}{4} + n,\;n} \right),\,\,\,n \in Z.\)