$\begin{array}{l} {\sin ^4}x + {\cos ^4}x = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x{\cos ^2}x\\ = 1 - 2{\left( {\sin x\cos x} \right)^2} = 1 - 2.{\left( {\dfrac{1}{2}\sin 2x} \right)^2} = 1 - 2.\dfrac{1}{4}{\sin ^2}2x\\ = 1 - \dfrac{1}{2}{\sin ^2}2x\\ 4\left( {{{\sin }^4}x + {{\cos }^4}x} \right) + \sqrt 3 \sin 4x = 2\\ \Leftrightarrow 4\left( {1 - \dfrac{1}{2}{{\sin }^2}2x} \right) + \sqrt 3 \sin 2x = 2\\ \Leftrightarrow 4 - 2{\sin ^2}2x + \sqrt 3 \sin 4x = 2\\ \Leftrightarrow 2 - 2{\sin ^2}2x + \sqrt 3 \sin 4x = 0\\ \Leftrightarrow 2 - 2.\dfrac{{1 - \cos 4x}}{2} + \sqrt 3 \sin 4x = 0\\ \Leftrightarrow \cos 4x + \sqrt 3 \sin 4x = - 1\\ \Leftrightarrow 2\cos \left( {4x - \dfrac{\pi }{3}} \right) = - 1\\ \Leftrightarrow \cos \left( {4x - \dfrac{\pi }{3}} \right) = - \dfrac{1}{2}\\ \Leftrightarrow \left[ \begin{array}{l} 4x - \dfrac{\pi }{3} = \dfrac{\pi }{2} + \dfrac{\pi }{6} + k2\pi \\ 4x - \dfrac{\pi }{3} = - \dfrac{\pi }{2} - \dfrac{\pi }{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} 4x = \dfrac{\pi }{3} + k2\pi \\ 4x = - \dfrac{\pi }{3} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{2}\\ x = - \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{2} \end{array} \right.\left( {k \in Z} \right) \end{array}$
