33В. а) Решите уравнение \(\frac{{2{{\sin }^2}x — \sin x}}{{{{\log }_2}\left( {\cos x} \right)}} = 0;\)
б) Найдите все корни принадлежащие промежутку \(\left[ { — 5\pi ; — \frac{{7\pi }}{2}} \right].\)
a) \(\frac{{2{{\sin }^2}x-\sin x}}{{{{\log }_2}\left( {\cos x} \right)}} = 0\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{2{{\sin }^2}x-\sin x = 0,}\\{\begin{array}{*{20}{c}}{{{\log }_2}\left( {\cos x} \right) \ne 0,}\\{\cos x > 0\;\;\;\;\;\;\;\;\;\;}\end{array}\,\,\,}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{\sin x\left( {2\sin x-1} \right) = 0,}\\{\begin{array}{*{20}{c}}{\cos x \ne 1,\;\;\;\;\;\;\;\;\;\;\;\;}\\{\cos x > 0\;\;\;\;\;\;\;\;\;\;\;\;}\end{array}\,\,\,\,\,}\end{array}} \right.\;\;\;\; \Leftrightarrow \) \( \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{\left[ {\begin{array}{*{20}{c}}{\sin x = 0,}\\{\sin x = \frac{1}{2}}\end{array}\,\,} \right.}\\{\begin{array}{*{20}{c}}{\cos x \ne 1,\;}\\{\cos x > 0\,\,}\end{array}}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left\{ {\begin{array}{*{20}{c}}{\left[ {\begin{array}{*{20}{c}}{x = 2\pi k,\,\,\,\,\;\;\;\;\;}\\{x = \pi + 2\pi k,\,\,\,}\\{x = \frac{\pi }{6} + 2\pi k,\;\,}\\{x = \frac{{5\pi }}{6} + 2\pi k,}\end{array}\,\,} \right.}\\{\begin{array}{*{20}{c}}{x \ne 2\pi n,\;}\\{\cos x > 0\,\,}\end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\;\;\;k,n \in Z\;\;\; \Leftrightarrow \;\;\;\;x = \frac{\pi }{6} + 2\pi k,\,\,\;\;k \in Z.\) \(x = \frac{\pi }{6} — 4\pi = — \frac{{23\pi }}{6}.\) Ответ: а) \(\frac{\pi }{6} + 2\pi k,\,\,\,\,\,k\, \in \,Z;\) б) \( — \frac{{23\pi }}{6}.\)
б) Отберём корни, принадлежащие отрезку \(\left[ { — 5\pi ; — \frac{{7\pi }}{2}} \right],\) с помощью тригонометрической окружности. Получим значение: