Задача 34. Решите уравнение: \(8{\sin ^4}x-10{\sin ^2}x + 3 = 0\)
ОТВЕТ: \(\,\dfrac{\pi }{4} + \dfrac{{\pi k}}{2};\quad \, \pm \dfrac{\pi }{3} + \pi k;\quad k \in Z.\)
\(8{\sin ^4}x-10{\sin ^2}x + 3 = 0.\) Пусть \({\sin ^2}x = t,\,\,\,\,\,t\, \in \,\left[ {0;1} \right].\) Тогда уравнение примет вид: \(8{t^2}-10t + 3 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = \dfrac{1}{2},}\\{t = \dfrac{3}{4}.}\end{array}} \right.\) Вернёмся к прежней переменной: \(\left[ {\begin{array}{*{20}{c}}{{{\sin }^2}x = \dfrac{1}{2},}\\{{{\sin }^2}x = \dfrac{3}{4}.}\end{array}} \right.\) Рассмотрим первое уравнение полученной совокупности: \({\sin ^2}x = \dfrac{1}{2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin x = \dfrac{{\sqrt 2 }}{2},\,\,}\\{\sin x = -\dfrac{{\sqrt 2 }}{2}}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{4} + 2\pi k,\,\,\,\,\,\,}\\{x = \dfrac{{3\pi }}{4} + 2\pi k,\,\,}\\{x = -\dfrac{\pi }{4} + 2\pi k,\,\,}\\{x = -\dfrac{{3\pi }}{4} + 2\pi k}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = \dfrac{\pi }{4} + \dfrac{{\pi k}}{2},\,\,\,\,k\, \in \,Z.\) Рассмотрим второе уравнение: \(\sin x = \dfrac{3}{4}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin x = \dfrac{{\sqrt 3 }}{2},}\\{\sin x = -\dfrac{{\sqrt 3 }}{2}}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{3} + 2\pi k,\,\,\,\,}\\{x = \dfrac{{2\pi }}{3} + 2\pi k,}\\{x = -\dfrac{\pi }{3} + 2\pi k,}\\{x = -\dfrac{{2\pi }}{3} + 2\pi k}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = \pm \dfrac{\pi }{3} + \pi k,\,\,\,\,\,\,\,k\, \in \,Z.\) Ответ: \(\,\dfrac{\pi }{4} + \dfrac{{\pi k}}{2};\quad \, \pm \dfrac{\pi }{3} + \pi k;\quad k \in Z.\)