Замена неизвестного t = sin x + cos x. Задача 7

ОТВЕТ: \(-\dfrac{\pi }{4} + {\left( {-1} \right)^k}\arcsin \dfrac{{\sqrt 2 }}{3} + \pi \,k;\;\,\,\;k \in Z.\)

\(\dfrac{{\sin x + \cos x-1}}{{\sin x + \cos x-2}} = \dfrac{{4\,\left( {\sin x + \cos x} \right)}}{{9-3\sin 2x}}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\dfrac{{\sin x + \cos x-1}}{{\sin x + \cos x-2}} = \dfrac{{4\,\left( {\sin x + \cos x} \right)}}{{9-3 \cdot 2 \cdot \sin x\cos x}}.\)

Пусть  \(\sin x + \cos x = t,\,\,\,\,t \in \left[ {-\sqrt 2 ;\sqrt 2 } \right].\)  Возведём обе части этой замены в квадрат:

\({\left( {\sin x + \cos x} \right)^2} = {t^2}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\sin ^2}x + 2\sin x\cos x + {\cos ^2}x = {t^2}\,\,\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,1 + 2\sin x\cos x = {t^2}\,\,\,\,\, \Leftrightarrow \,\,\,\,\,2\sin x\cos x = {t^2}-1.\)

Тогда исходное уравнение примет вид:

\(\dfrac{{t-1}}{{t-2}} = \dfrac{{4t}}{{9-3\left( {{t^2}-1} \right)}}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\dfrac{{t-1}}{{t-2}} = \dfrac{{4t}}{{12-3{t^2}}}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\dfrac{{t-1}}{{t-2}} + \dfrac{{4t}}{{3\left( {t-2} \right)\left( {t + 2} \right)}} = 0\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\dfrac{{3{t^2} + 7t-6}}{{3\left( {t-2} \right)\left( {t + 2} \right)}} = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{3{t^2} + 7t-6 = 0,}\\{t \ne  \pm 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\left[ {\begin{array}{*{20}{c}}{t = \dfrac{2}{3},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{t = -3\, \notin \left[ {-\sqrt 2 ;\sqrt 2 } \right],}\end{array}} \right.}\\{t \ne  \pm 2.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\)

Вернёмся к прежней переменной:

\(\sin x + \cos x = \dfrac{2}{3}.\)

Получили уравнение вида:  \(a\sin x + b\cos x = c,\;\;\)где \(\;a = 1,\,\,\,\,b = 1,\;\;c = \dfrac{2}{3}.\)

Разделим обе части последнего уравнения на \(\sqrt {{a^2} + {b^2}} ,\;\) то есть на  \(\sqrt {{1^2} + {1^2}}  = \sqrt 2 .\)

\(\dfrac{1}{{\sqrt 2 }}\sin x + \dfrac{1}{{\sqrt 2 }}\cos x = \dfrac{2}{{3\sqrt 2 }}\;\;\;\; \Leftrightarrow \;\;\;\;\cos \dfrac{\pi }{4}\sin x + \sin \dfrac{\pi }{4}\cos x = \dfrac{{\sqrt 2 }}{3}\;\;\;\; \Leftrightarrow \)

\( \Leftrightarrow \;\;\;\;\sin \left( {x + \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{3}\;\;\;\; \Leftrightarrow \;\;\;\;x + \dfrac{\pi }{4} = {\left( {-1} \right)^k}\arcsin \dfrac{{\sqrt 2 }}{3} + \pi k\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x = {\left( {-1} \right)^k}\arcsin \dfrac{{\sqrt 2 }}{3}-\dfrac{\pi }{4} + \pi k,\,\,\,\,k \in Z.\)

Ответ:  \(-\dfrac{{\rm{\pi }}}{4} + {\left( {-1} \right)^k}\arcsin \dfrac{{\sqrt 2 }}{3} + \pi \,k;\;\,\,\;k \in Z.\)