Задача 8. Решите уравнение: \(\sin \,\left( {x + \dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}\,\left( {1-\sin x\cos x} \right)\)
ОТВЕТ: \(2\pi \,k;\;\,\,\,\,\dfrac{\pi }{2} + 2\pi \,k;\;\,\,\;k \in Z.\)
\(\sin \,\left( {x + \dfrac{{\rm{\pi }}}{4}} \right) = \dfrac{1}{{\sqrt 2 }}\,\left( {1-\sin x\cos x} \right).\) Воспользуемся формулой суммы синуса: \(\sin \left( {\alpha + \beta } \right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta .\) \(\sin \,\left( {x + \dfrac{{\rm{\pi }}}{4}} \right) = \dfrac{1}{{\sqrt 2 }}\,\left( {1-\sin x\cos x} \right)\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\sin x\cos \dfrac{\pi }{4} + \cos x\sin \dfrac{\pi }{4} = \dfrac{{\sqrt 2 }}{2}\,\left( {1-\sin x\cos x} \right)\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\dfrac{{\sqrt 2 }}{2}\sin x + \dfrac{{\sqrt 2 }}{2}\cos x = \dfrac{{\sqrt 2 }}{2}\,\left( {1-\sin x\cos x} \right)\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\sin x + \cos x = 1-\sin x\cos x.\) Пусть \(\sin x + \cos x = t,\,\,\,\,t \in \left[ {-\sqrt 2 ;\sqrt 2 } \right].\) Возведём обе части этой замены в квадрат: \({\left( {\sin x + \cos x} \right)^2} = {t^2}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\sin ^2}x + 2\sin x\cos x + {\cos ^2}x = {t^2}\,\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,1 + 2\sin x\cos x = {t^2}\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\sin x\cos x = \dfrac{{{t^2}-1}}{2}.\) Тогда последнее уравнение примет вид: \(t = 1-\dfrac{{{t^2}-1}}{2}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{t^2} + 2t-3 = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{t = -3 \notin \left[ {-\sqrt 2 ;\sqrt 2 } \right].}\end{array}} \right.\) Вернёмся к прежней переменной: \(\sin x + \cos x = 1.\) Получили уравнение вида: \(a\sin x + b\cos x = c,\;\;\)где \(\;a = 1,\,\,\,\,b = 1,\;\;c = 1.\) Разделим обе части последнего уравнения на \(\sqrt {{a^2} + {b^2}} ,\;\) то есть на \(\sqrt {{1^2} + {1^2}} = \sqrt 2 .\) \(\dfrac{1}{{\sqrt 2 }}\sin x + \dfrac{1}{{\sqrt 2 }}\cos x = \dfrac{2}{{3\sqrt 2 }}\;\;\;\; \Leftrightarrow \;\;\;\;\cos \dfrac{\pi }{4}\sin x + \sin \dfrac{\pi }{4}\cos x = \dfrac{{\sqrt 2 }}{2}\;\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\;\sin \left( {x + \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{2}\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x + \dfrac{\pi }{4} = \dfrac{\pi }{4} + 2\pi k,\,\;\;\;\,\,\,}\\{x + \dfrac{\pi }{4} = \dfrac{{3\pi }}{4} + 2\pi k\,\,\,\,\,\,\,}\end{array} \Leftrightarrow \;\;\;\;} \right.\left[ {\begin{array}{*{20}{c}}{x = 2\pi k,\,\;\;\;\;\;\;\;\;\,\,\,}\\{x = \dfrac{\pi }{2} + 2\pi k,\;\,\,\,\,\,\,}\end{array}k\, \in \,Z.} \right.\) Ответ: \(2\pi \,k;\;\,\,\,\,\dfrac{\pi }{2} + 2\pi \,k;\;\,\,\;k \in Z.\)