$\begin{array}{l} 1)2{\sin ^2}x + \sqrt 3 \sin 2x = 3\\ \Leftrightarrow 2{\sin ^2}x - 1 + \sqrt 3 \sin 2x = 2\\ \Leftrightarrow \sqrt 3 \sin 2x - \cos 2x = 2\\ \Leftrightarrow 2\left( {\dfrac{{\sqrt 3 }}{2}\sin 2x - \dfrac{1}{2}\cos 2x} \right) = 2\\ \Leftrightarrow 2\sin \left( {2x - \dfrac{\pi }{6}} \right) = 2\\ \Leftrightarrow \sin \left( {2x - \dfrac{\pi }{6}} \right) = 1\\ \Rightarrow 2x - \dfrac{\pi }{6} = \dfrac{\pi }{2} + k2\pi \\ \Leftrightarrow 2x = \dfrac{{2\pi }}{3} + k2\pi \\ \Leftrightarrow x = \dfrac{\pi }{3} + k2\pi \left( {k \in \mathbb{Z}} \right)\\ 3)8\cos x = \dfrac{{\sqrt 3 }}{{\sin x}} + \dfrac{1}{{\cos x}}\\ \Leftrightarrow 8\cos x = \dfrac{{\sqrt 3 \cos x + \sin x}}{{\sin x\cos x}}\\ \Leftrightarrow 8\cos {x^2}\sin x = \sqrt 3 \cos x + \sin x\\ \Leftrightarrow 4\left( {1 + \cos 2x} \right)\sin x = \sqrt 3 \cos x + \sin x\\ \Leftrightarrow 4\sin x + 4\cos 2x\sin x = \sqrt 3 \cos x + \sin x\\ \Leftrightarrow 4\sin x + 2\sin 3x - 2\sin x = \sqrt 3 \cos x + \sin x\\ \Leftrightarrow 2\sin 3x = \sqrt 3 \cos x - \sin x\\ \Leftrightarrow 2\sin 3x = 2\left( {\dfrac{{\sqrt 3 }}{2}\cos x - \dfrac{1}{2}\sin x} \right)\\ \Leftrightarrow 2\sin 3x = 2\left( {\sin \dfrac{\pi }{3}\cos x - \cos \dfrac{\pi }{3}\sin x} \right)\\ \Leftrightarrow 2\sin 3x = 2\sin \left( {\dfrac{\pi }{3} - x} \right)\\ \Leftrightarrow \left[ \begin{array}{l} 3x = \dfrac{\pi }{3} - x + k2\pi \\ 3x = \dfrac{{2\pi }}{3} + x + l2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{2}\\ x = \dfrac{\pi }{3} + l\pi \end{array} \right.\left( {k,l \in \mathbb{Z}} \right)\\ 5)\sin 5x + \cos 5x = \sqrt 2 \cos 13x\\ \Leftrightarrow \sqrt 2 \left( {\dfrac{{\sqrt 2 }}{2}\cos 5x + \dfrac{{\sqrt 2 }}{2}\sin 5x} \right) = \sqrt 2 \cos 13x\\ \Leftrightarrow \cos \left( {5x - \dfrac{\pi }{4}} \right) = \cos 13x\\ \Leftrightarrow \left[ \begin{array}{l} 5x - \dfrac{\pi }{4} = 13x + k2\pi \\ 5x - \dfrac{\pi }{4} = - 13x + l2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} - 8x = \dfrac{\pi }{4} + k2\pi \\ 18x = \dfrac{\pi }{4} + l2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{{ - \pi }}{{32}} + \dfrac{{k\pi }}{4}\\ x = \dfrac{\pi }{{72}} + \dfrac{{l2\pi }}{9} \end{array} \right.\left( {k,l \in \mathbb{Z}} \right) \end{array}$
