Задача 2. Решите уравнение: \(2\sin x\cos x-\sin x + \cos x = -1\)
ОТВЕТ: \(\dfrac{\pi }{2} + 2\pi k;\;\,\,\;\pi + 2\pi k;\;\,\,\;k \in Z.\)
\(2\sin x\cos x-\sin x + \cos x = -1\,\,\,\,\, \Leftrightarrow \,\,\,\,\,2\sin x\cos x-\left( {\sin x-\cos x} \right) = -1.\) Пусть \(\sin x-\cos x = t,\,\,\,\,t \in \left[ {-\sqrt 2 ;\sqrt 2 } \right].\) Возведём обе части этой замены в квадрат: \({\left( {\sin x-\cos x} \right)^2} = {t^2}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\sin ^2}x-2\sin x\cos x + {\cos ^2}x = {t^2}\,\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,1-2\sin x\cos x = {t^2}\,\,\,\,\, \Leftrightarrow \,\,\,\,\,2\sin x\cos x = 1-{t^2}.\) Тогда исходное уравнение примет вид: \(1-{t^2}-t = -1\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{t^2} + t-2 = 0\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{t = -2 \notin \left[ {-\sqrt 2 ;\sqrt 2 } \right].}\end{array}} \right.\) 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 \(\sqrt {{1^2} + {{\left( {-1} \right)}^2}} = \sqrt 2 .\) \(\dfrac{1}{{\sqrt 2 }}\sin x-\dfrac{1}{{\sqrt 2 }}\cos x = \dfrac{1}{{\sqrt 2 }}\;\;\;\; \Leftrightarrow \;\;\;\;\cos \dfrac{\pi }{4}\sin x-\sin \dfrac{\pi }{4}\cos x = \dfrac{1}{{\sqrt 2 }}\;\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\;\sin \left( {x-\dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{2}\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{x-\dfrac{\pi }{4} = \dfrac{\pi }{4} + 2\pi k,\,\;\;\;\,\,\,}\\{x-\dfrac{\pi }{4} = \dfrac{{3\pi }}{4} + 2\pi k\,\,\,\,\,\,\,}\end{array} \Leftrightarrow \;\;\;\;} \right.\left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{2} + 2\pi k,\,\;\;\,}\\{x = \pi + 2\pi k,\;\,\,\,\,}\end{array}k\, \in \,Z.} \right.\) Ответ: \(\dfrac{\pi }{2} + 2\pi k;\;\,\,\;\pi + 2\pi k;\;\,\,\;k \in Z.\)