Задача 6. Решите уравнение: \({\sin ^3}x-{\cos ^3}x = \sin 2x-1\)
ОТВЕТ: \(\dfrac{\pi }{4} + \pi k;\;\,\,\,\;2\pi k;\,\,\,\,\,\,-\dfrac{\pi }{2} + 2\pi k;\;\;\,\,k \in Z.\)
\({\sin ^3}x-{\cos ^3}x = \sin 2x-1\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\left( {\sin x-\cos x} \right)\left( {{{\sin }^2}x + \sin x\cos x + {{\cos }^2}x} \right) + {\sin ^2}x-2\sin x\cos x + {\cos ^2}x = 0\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\;\left( {\sin x-\cos x} \right)\left( {1 + \sin x\cos x} \right) + {\left( {\sin x-\cos x} \right)^2} = 0\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\;\left( {\sin x-\cos x} \right)\left( {1 + \sin x\cos x + \sin x-\cos x} \right) = 0\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\;\left[ {\begin{array}{*{20}{c}}{\sin x-\cos x = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{1 + \sin x\cos x + \sin x-\cos x = 0.}\end{array}} \right.\) Первое уравнение последней совокупности \(\sin x-\cos x = 0\) является однородным тригонометрическим уравнением первой степени: \(\sin x-\cos x = 0\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\;{\rm{tg}}\,x-1 = 0\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\;{\rm{tg}}\,x = 1\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\;x = \dfrac{\pi }{4} + \pi k,\,\,\,\,\,k \in Z.\) Рассмотрим второе уравнение: \(1 + \sin x\cos x + \sin x-\cos x = 0.\) Пусть \(\sin x-\cos x = t,\,\,\,\,t \in \left[ {-\sqrt 2 ;\sqrt 2 } \right].\) Возведём обе части этой замены в квадрат: \({\left( {\sin x-\cos x} \right)^2} = {t^2}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\sin ^2}x-2\sin x\cos x + {\cos ^2}x = {t^2}\,\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,1-2\sin x\cos x = {t^2}\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\sin x\cos x = \dfrac{{1-{t^2}}}{2}.\) Тогда последнего уравнение примет вид: \(1 + \dfrac{{1-{t^2}}}{2} + t = 0\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{t^2}-2t-3 = 0\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = -1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{t = 3 \notin \left[ {-\sqrt 2 ;\sqrt 2 } \right].}\end{array}} \right.\) 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 исходного уравнения на \(\sqrt {{a^2} + {b^2}} ,\;\) то есть на \(\sqrt {{1^2} + {{\left( {-1} \right)}^2}} = \sqrt 2 .\) \(\dfrac{1}{{\sqrt 2 }}\sin x-\dfrac{1}{{\sqrt 2 }}\cos x = -\dfrac{1}{{\sqrt 2 }}\;\;\;\; \Leftrightarrow \;\;\;\;\cos \dfrac{\pi }{4}\sin x-\sin \dfrac{\pi }{4}\cos x = -\dfrac{1}{{\sqrt 2 }}\;\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\;\sin \left( {x-\dfrac{\pi }{4}} \right) = -\dfrac{{\sqrt 2 }}{2}\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x-\dfrac{\pi }{4} = -\dfrac{\pi }{4} + 2\pi k,\,\;\;\;\,\,\,}\\{x-\dfrac{\pi }{4} = -\dfrac{{3\pi }}{4} + 2\pi k\,\,\,\,\,\,\,}\end{array} \Leftrightarrow \;\;\;\;} \right.\left[ {\begin{array}{*{20}{c}}{x = 2\pi k,\;\,\,\,\,\,\,\,\,\,\,\;\;\;\;\,\,\,}\\{x = -\dfrac{\pi }{2} + 2\pi k,\;\,\,\,\,\,}\end{array}k\, \in \,Z.} \right.\) Ответ: \(\dfrac{\pi }{4} + \pi k;\;\,\,\,\;2\pi k;\,\,\,\,\,\,-\dfrac{\pi }{2} + 2\pi k;\;\;\,\,k \in Z.\)