Уравнения, сводящиеся к простейшим заменой неизвестного. Задача 39

ОТВЕТ: \(\dfrac{\pi }{2} + \pi k;\,\,\,\,\,\dfrac{\pi }{6} + 2\pi k;\quad \dfrac{{5\pi }}{6} + 2\pi k;\quad k \in Z.\)

\(2{\sin ^3}x-{\sin ^2}x-2\sin x + 1 = 0.\)

Пусть \(\sin x = t,\,\,\,\,t\, \in \,\left[ {-1;1} \right]\). Тогда уравнение примет вид:

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

\( \Leftrightarrow \,\,\,\,\,\,\,\left( {2t-1} \right)\left( {{t^2}-1} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{2t-1 = 0,}\\{{t^2}-1 = 0\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = \dfrac{1}{2},\,\,}\\{t =  \pm 1.}\end{array}} \right.\)

Вернёмся к прежней переменной:

\(\left[ {\begin{array}{*{20}{c}}{\sin x = \dfrac{1}{2},}\\{\sin x =  \pm 1}\end{array}} \right.\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{6} + 2\pi k,\,\,}\\{x = \dfrac{{5\pi }}{6} + 2\pi k,}\\{x = \dfrac{\pi }{2} + \pi k,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,k\, \in \,Z.\)

Ответ:  \(\dfrac{\pi }{2} + \pi k;\,\,\,\,\,\dfrac{\pi }{6} + 2\pi k;\quad \dfrac{{5\pi }}{6} + 2\pi k;\quad k \in Z.\)