Задача 1. Решите уравнение: \(2{\cos ^2}x = 3\sin x\)
ОТВЕТ: \(\dfrac{\pi }{6} + 2\pi k;\quad \,\dfrac{{5\pi }}{6} + 2\pi k;\quad k \in Z.\)
\(2{\cos ^2}x = 3\sin x\) Так как \({\sin ^2}x + {\cos ^2}x = 1\), то \({\cos ^2}x = 1-{\sin ^2}x\). Тогда уравнение примет вид: \(2\left( {1-{{\sin }^2}x} \right) = 3\sin x\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,2{\sin ^2}x + 3\sin x-2 = 0.\) Пусть \(\sin x = t,\,\,\,\,t\, \in \,\left[ {-1;1} \right]\). Тогда: \(2{t^2} + 3t-2 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = \dfrac{1}{2},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{t = -2\,\,\, \notin \,\left[ {-1;1} \right].}\end{array}} \right.\) Вернёмся к прежней переменной: \(\sin x = \dfrac{1}{2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{6} + 2\pi k,\,\,\,}\\{x = \dfrac{{5\pi }}{6} + 2\pi k,}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,\,k\, \in \,Z.\) Ответ: \(\dfrac{\pi }{6} + 2\pi k;\quad \,\dfrac{{5\pi }}{6} + 2\pi k;\quad k \in Z.\)