а) \(1-\cos 2x-\sqrt 3 \sin \left( {x + {\rm{\pi }}} \right) = \sqrt 3 + 2\sin x\)
Воспользуемся формулой приведения \(\sin \left( {x + {\rm{\pi }}} \right) = -\sin x\) и формулой косинуса двойного угла: \(\cos 2x = 1-2{\sin ^2}x:\)
\(1-\cos 2x-\sqrt 3 \sin \left( {x + {\rm{\pi }}} \right) = \sqrt 3 + 2\sin x\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,1-\left( {1-2{{\sin }^2}x} \right) + \sqrt 3 \sin x = \sqrt 3 + 2\sin x\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,2{\sin ^2}x-2\sin x + \sqrt 3 \sin x-\sqrt 3 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,2\sin x\left( {\sin x-1} \right) + \sqrt 3 \left( {\sin x-1} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\left( {\sin x-1} \right)\left( {2\sin x + \sqrt 3 } \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left[ \begin{array}{l}\sin x-1 = 0,\\2\sin x + \sqrt 3 = 0\end{array} \right.\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\left[ \begin{array}{l}\sin x = 1,\\\sin x = -\dfrac{{\sqrt 3 }}{2}\end{array} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left[ \begin{array}{l}x = \dfrac{{\rm{\pi }}}{2} + 2{\rm{\pi }}k,\\x = -\dfrac{{\rm{\pi }}}{3} + 2{\rm{\pi }}k,\\x = -\dfrac{{{\rm{2\pi }}}}{3} + 2{\rm{\pi }}k\end{array} \right.\,\,\,\,\,\,\,\,k \in Z.\)
б) Отберём корни, принадлежащие отрезку \(\left[ {-5{\rm{\pi }};\,-\dfrac{{7{\rm{\pi }}}}{2}} \right]\) с помощью тригонометрической окружности. Получим значения:
\(x = \dfrac{{\rm{\pi }}}{2}-4{\rm{\pi }} = -\dfrac{{{\rm{7\pi }}}}{2};\,\,\;\,x = -\dfrac{{\rm{\pi }}}{3}-4{\rm{\pi }} = -\dfrac{{{\rm{13\pi }}}}{3};\)
\(x = -\dfrac{{{\rm{2\pi }}}}{3}-4{\rm{\pi }} = -\dfrac{{{\rm{14\pi }}}}{3}.\)
Ответ: а) \(\dfrac{{\rm{\pi }}}{2} + 2{\rm{\pi }}k;\quad -\dfrac{{\rm{\pi }}}{3} + 2{\rm{\pi }}k;\quad -\dfrac{{{\rm{2\pi }}}}{3} + 2{\rm{\pi }}k;\,\,\,\,\,\,\,\,k \in Z;\)
б) \(-\dfrac{{{\rm{14\pi }}}}{3};\;\;-\dfrac{{{\rm{13\pi }}}}{3};\;\;-\dfrac{{{\rm{7\pi }}}}{2}.\)