Задача 1. Решите систему уравнений \(\left\{ {\begin{array}{*{20}{c}} {\sin x\,\sin y = \dfrac{1}{4}} \\ {\cos x\,\cos y = \dfrac{3}{4}} \end{array}} \right.\)
Ответ
ОТВЕТ: \(\left( {-\dfrac{\pi }{6} + \pi \left( {n + k} \right),-\dfrac{\pi }{6} + \pi \left( {n-k} \right)} \right),\;\;\left( {\dfrac{\pi }{6} + \pi \left( {n + k} \right),\dfrac{\pi }{6} + \pi \left( {n-k} \right)} \right),\,\,n,k \in Z.\)
Решение
\(\left\{ {\begin{array}{*{20}{c}}{\sin x\,\sin y = \dfrac{1}{4},}\\{\cos x\,\cos y = \dfrac{3}{4}}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\cos x\cos y-\sin x\sin y = \dfrac{1}{2},}\\{\cos x\cos y + \sin x\sin y = 1}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\cos \left( {x + y} \right) = \dfrac{1}{2},}\\{\cos \left( {x-y} \right) = 1\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x + y = \dfrac{\pi }{3} + 2\pi n,}\\{x-y = 2\pi k,\,\,\,\,\,\,\,\,\,\,\,}\end{array}\,\,\,} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x + y = -\dfrac{\pi }{3} + 2\pi n,}\\{x-y = 2\pi k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{2x = \dfrac{\pi }{3} + 2\pi k + 2\pi n,}\\{2y = \dfrac{\pi }{3} + 2\pi n-2\pi k,}\end{array}\,\,\,} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{2x = -\dfrac{\pi }{3} + 2\pi k + 2\pi n,}\\{2y = -\dfrac{\pi }{3} + 2\pi n-2\pi k}\end{array}} \right.}\end{array}} \right.\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{6} + \pi \left( {k + n} \right),}\\{y = \dfrac{\pi }{6} + \pi \left( {n-k} \right),}\end{array}\,\,\,\,} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x = -\dfrac{\pi }{6} + \pi \left( {k + n} \right),}\\{y = -\dfrac{\pi }{6} + \pi \left( {n-k} \right),}\end{array}} \right.}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,n,\,k\, \in \,Z.\)
Ответ: \(\left( {-\dfrac{{\rm{\pi }}}{6} + {\rm{\pi }}\left( {n + k} \right),-\dfrac{{\rm{\pi }}}{6} + {\rm{\pi }}\left( {n-k} \right)} \right),\;\;\left( {\dfrac{{\rm{\pi }}}{6} + {\rm{\pi }}\left( {n + k} \right),\dfrac{{\rm{\pi }}}{6} + {\rm{\pi }}\left( {n-k} \right)} \right),\,\,n,k \in Z.\)