а) Воспользуемся формулой синуса суммы: \(\sin \left( {{\rm{\alpha }} + {\rm{\beta }}} \right) = \sin {\rm{\alpha }}\cos {\rm{\beta }} + \cos {\rm{\alpha }}\sin {\rm{\beta }}{\rm{.}}\)
\(2\sqrt 2 \sin \left( {x + \dfrac{{\rm{\pi }}}{3}} \right) + 2{\cos ^2}x = 2 + \sqrt 6 \cos x\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,2\sqrt 2 \left( {\sin x\cos \dfrac{{\rm{\pi }}}{3} + \cos x\sin \dfrac{{\rm{\pi }}}{3}} \right) + 2{\cos ^2}x = 2 + \sqrt 6 \cos x\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,2\sqrt 2 \cdot \dfrac{1}{2}\sin x + 2\sqrt 2 \cdot \dfrac{{\sqrt 3 }}{2}\cos x + 2{\cos ^2}x = 2 + \sqrt 6 \cos x\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\sqrt 2 \sin x + \sqrt 6 \cos x + 2{\cos ^2}x-2-\sqrt 6 \cos x = 0\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\sqrt 2 \sin x + 2\left( {1-{{\sin }^2}x} \right)-2 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\sqrt 2 \sin x + 2-2{\sin ^2}x-2 = 0\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\sin x\left( {\sqrt 2 -2\sin x} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left[ \begin{array}{l}\sin x = 0,\\\sqrt 2 -2\sin x = 0\end{array} \right.\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\left[ \begin{array}{l}\sin x = 0,\\\sin x = \dfrac{{\sqrt 2 }}{2}\end{array} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ \begin{array}{l}x = {\rm{\pi }}k,\\x = \dfrac{{\rm{\pi }}}{4} + 2{\rm{\pi }}k,\\x = \dfrac{{{\rm{3\pi }}}}{4} + 2{\rm{\pi }}k,\end{array} \right.\,\,\,\,\,\,k \in Z.\)
б) Отберём корни, принадлежащие отрезку \(\left[ {-3{\rm{\pi ;}}\,-\dfrac{{{\rm{3\pi }}}}{2}} \right]\) с помощью тригонометрической окружности. Получим значения:
\(x = -3{\rm{\pi }};\;\;\;x = -2{\rm{\pi }};\,\,\,\,\,x = \dfrac{{\rm{\pi }}}{4}-2{\rm{\pi }} = -\dfrac{{{\rm{7\pi }}}}{4}.\)
Ответ: а) \({\rm{\pi }}k;\quad \dfrac{{\rm{\pi }}}{4} + 2{\rm{\pi }}k;\,\,\,\,\,\,\dfrac{{{\rm{3\pi }}}}{4} + 2{\rm{\pi }}k;\,\,\,\,\,\,\,k \in Z;\)
б) \(-3{\rm{\pi }};\,\,\,\,-2{\rm{\pi ;}}\,\,\,\,\,-\dfrac{{{\rm{7\pi }}}}{4}.\)