\(5{\sin ^2}x + \sqrt 3 \sin x\cos x + 6{\cos ^2}x = 5.\)
Так как \({\sin ^2}x + {\cos ^2}x = 1,\) то \(5 = 5 \cdot 1 = 5\left( {{{\sin }^2}x + {{\cos }^2}x} \right).\) Тогда уравнение примет вид:
\(5{\sin ^2}x + \sqrt 3 \sin x\cos x + 6{\cos ^2}x = 5\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\sqrt 3 \sin x\cos x + {\cos ^2}x = 0\;\;\;\; \Leftrightarrow \;\;\;\;\cos x\left( {\sqrt 3 \sin x + \cos x} \right) = 0\;\;\;\; \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\cos x = 0,\;\;\;\;\;\;}\\{\sqrt 3 {\rm{tg}}\,x + 1 = 0}\end{array}} \right.\;\;\;\; \Leftrightarrow \)
\( \Leftrightarrow \;\;\;\;\left[ {\begin{array}{*{20}{c}}{\cos x = 0,\;\;\;\;\;\;}\\{{\rm{tg}}\,x = -\dfrac{{\sqrt 3 }}{3}\;\;\;}\end{array}} \right.\;\;\;\; \Leftrightarrow \;\;\;\;\;\left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{2} + \pi k,\;\;\,}\\{x = -\dfrac{\pi }{6} + \pi k,}\end{array}} \right.\;\;\;\;k \in Z.\)
Ответ: \(-\dfrac{\pi }{6} + \pi k;\;\;\dfrac{\pi }{2} + \pi k;\;\;k \in Z.\)