a) \(\sin x\cos x = \dfrac{1}{4}\;\;\;\; \Leftrightarrow \;\;\;\;2\sin x\cos x = \dfrac{1}{4} \cdot 2.\)
Так как \(2\sin x\cos x = \sin 2x\), то уравнение примет вид:
\(\sin 2x = \dfrac{1}{2}\;\;\;\, \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{2x = \dfrac{\pi }{6} + 2\pi k,\,\;\;\;\,\,\,}\\{2x = \dfrac{{5\pi }}{6} + 2\pi k\,\,\,\,\,\,\,}\end{array} \Leftrightarrow } \right.\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{{12}} + \pi k,\,\;\,\,\,\;}\\{x = \dfrac{{5\pi }}{{12}} + \pi k,\,\,\,\,\,\,}\end{array}k\, \in \,Z.} \right.\,\)
б) Отберём корни, принадлежащие отрезку \(\left[ { — \pi ;\pi } \right],\) с помощью тригонометрической окружности. Получим значения:
\(x = \dfrac{\pi }{{12}} — \pi = — \dfrac{{11\pi }}{{12}};\) \(x = \dfrac{{5\pi }}{{12}} — \pi = — \dfrac{{7\pi }}{{12}};\,\,\,\,\,\;x = \dfrac{\pi }{{12}};\,\;\,\,\,\,x = \dfrac{{5\pi }}{{12}}.\)
Ответ: а) \(\dfrac{\pi }{{12}} + \pi k,\,\,\,\,\,\,\,\dfrac{{5\pi }}{{12}} + \pi k,\,\,\,\,\,\,\,k\, \in \,Z;\)
б) \( — \dfrac{{11\pi }}{{12}};\,\,\,\, — \dfrac{{7\pi }}{{12}};\,\,\;\dfrac{\pi }{{12}};\,\;\,\,\dfrac{{5\pi }}{{12}}.\)