Уравнения, сводящиеся к простейшим заменой неизвестного. Задача 40

ОТВЕТ: \( \pm \dfrac{\pi }{3} + 2\pi k;\quad k \in Z.\)

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

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

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

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

Вернёмся к прежней переменной:

\(\cos x = \dfrac{1}{2}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x =  \pm \dfrac{\pi }{3} + 2\pi k,\,\,\,\,\,\,k\, \in \,Z.\)

Ответ:   \( \pm \dfrac{\pi }{3} + 2\pi k,\,\,\,\,\,\,k\, \in \,Z.\)