\(3{\sin ^2}\dfrac{x}{3} + 4{\cos ^2}\dfrac{x}{3} = 3 + \sqrt 3 \sin \dfrac{x}{3}\cos \dfrac{x}{3}.\)
Так как \({\sin ^2}\dfrac{x}{3} + {\cos ^2}\dfrac{x}{3} = 1\), то \(3 = 3 \cdot 1 = 3\left( {{{\sin }^2}\dfrac{x}{3} + {{\cos }^2}\dfrac{x}{3}} \right).\)
\(3{\sin ^2}\dfrac{x}{3} + 4{\cos ^2}\dfrac{x}{3} = 3{\sin ^2}\dfrac{x}{3} + 3{\cos ^2}\dfrac{x}{3} + \sqrt 3 \sin \dfrac{x}{3}\cos \dfrac{x}{3}\;\;\;\; \Leftrightarrow \)
\( \Leftrightarrow \;\;\;\;\sqrt 3 \sin \dfrac{x}{3}\cos \dfrac{x}{3}-{\cos ^2}\dfrac{x}{3} = 0\;\;\;\; \Leftrightarrow \;\;\;\;\cos \dfrac{x}{3}\left( {\sqrt 3 \sin \dfrac{x}{3}-\cos \dfrac{x}{3}} \right) = 0\;\;\;\; \Leftrightarrow \)
\( \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\cos \dfrac{x}{3} = 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\{\sqrt 3 \sin \dfrac{x}{3}-\cos \dfrac{x}{3} = 0}\end{array}\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\cos \dfrac{x}{3} = 0,\;\;\;\;\;\;}\\{\sqrt 3 {\rm{tg}}\dfrac{x}{3}-1 = 0}\end{array}} \right.} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\cos \dfrac{x}{3} = 0,\;}\\{{\rm{tg}}\dfrac{x}{3} = \dfrac{1}{{\sqrt 3 }}}\end{array}} \right.\;\;\;\; \Leftrightarrow \)
\( \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\dfrac{x}{3} = \dfrac{\pi }{2} + \pi k,\,\,}\\{\dfrac{x}{3} = \dfrac{\pi }{6} + \pi k\,\;}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\;\left[ {\begin{array}{*{20}{c}}{x = \dfrac{{3\pi }}{2} + 3\pi k,\,\,}\\{x = \dfrac{\pi }{2} + 3\pi k,\,\,\,\,\,}\end{array}\;\;\;\;k\, \in \,Z.} \right.\,\)
Ответ: \(\dfrac{\pi }{2} + 3\pi k;\quad \dfrac{{3\pi }}{2} + 3\pi k;\quad k \in Z.\)