Уравнения, сводящиеся к простейшим заменой неизвестного. Задача 3

Задача 3. Решите уравнение: \(\sqrt 2 {\sin ^2}x + \sin x = 0\)

Ответ

ОТВЕТ: \(\pi k;\,\,\,\,\,-\dfrac{\pi }{4} + 2\pi k;\quad \,-\dfrac{{3\pi }}{4} + 2\pi k;\quad k \in Z.\)

Решение

\(\sqrt 2 {\sin ^2}x + \sin x = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\sin x\left( {\sqrt 2 \sin x + 1} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin x = 0,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\sqrt 2 \sin x + 1 = 0}\end{array}\,\,\,\,\,\,\, \Leftrightarrow } \right.\)

\( \Leftrightarrow \,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin x = 0,}\\{\sin x = -\dfrac{{\sqrt 2 }}{2}}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \pi k,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x = -\dfrac{\pi }{4} + 2\pi k,\,\,\,}\\{x = -\dfrac{{3\pi }}{4} + 2\pi k,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,k\, \in \,Z.\)

Ответ: \(\pi k;\,\,\,\,\,-\dfrac{\pi }{4} + 2\pi k;\quad \,-\dfrac{{3\pi }}{4} + 2\pi k;\quad k \in Z.\)