\(\sin x-\dfrac{1}{{\sin x-1}} = \dfrac{5}{2}.\)
Пусть \(\sin x = t,\,\,\,\,t\, \in \,\left[ {-1;1} \right]\). Тогда уравнение примет вид:
\(t-\dfrac{1}{{t-1}} = \dfrac{5}{2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{2{t^2}-7t + 3 = 0,}\\{t \ne 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = \dfrac{1}{2},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{t = 3\,\, \notin \,\left[ {-1;1} \right].}\end{array}} \right.\)
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: \(\dfrac{\pi }{6} + 2\pi k;\quad \,\dfrac{{5\pi }}{6} + 2\pi k;\quad k \in Z.\)