Đáp án:

$\begin{array}{l}
1)4{\sin ^2}x - 2\left( {\sqrt 3  + 1} \right)\sin x + \sqrt 3  = 0\\
 \Rightarrow 4{\sin ^2}x - 2\sqrt 3 \sin x - 2\sin x + \sqrt 3  = 0\\
 \Rightarrow \left( {2{\mathop{\rm sinx}\nolimits}  - \sqrt 3 } \right)\left( {2\sin x - 1} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
\sin x = \dfrac{{\sqrt 3 }}{2}\\
\sin x = \dfrac{1}{2}
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{3} + k2\pi \\
x = \dfrac{{2\pi }}{3} + k2\pi \\
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\\
2)\dfrac{1}{{{{\sin }^2}x}} + 3{\tan ^2}x = 5\\
\left( {dkxd:\left\{ \begin{array}{l}
\cos x \ne 0\\
\sin x \ne 0
\end{array} \right. \Rightarrow x \ne \dfrac{{k\pi }}{2}} \right)\\
 \Rightarrow {\cot ^2}x + 1 + 3{\tan ^2}x = 5\\
 \Rightarrow {\cot ^2}x + \dfrac{3}{{{{\cot }^2}x}} = 4\\
 \Rightarrow {\cot ^4}x - 4{\cot ^2}x + 3 = 0\\
 \Rightarrow \left( {{{\cot }^2}x - 1} \right)\left( {{{\cot }^2}x - 3} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
{\cot ^2}x = 1\\
{\cot ^2}x = 3
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x =  \pm \dfrac{\pi }{4} + k\pi \left( {tm} \right)\\
x =  \pm \dfrac{\pi }{6} + k\pi \left( {tm} \right)
\end{array} \right.\\
3)4{\cos ^3}x + 3\sqrt 2 \sin 2x = 8\cos x\\
 \Rightarrow 4{\cos ^3}x + 6\sqrt 2 .cosx.sinx - 8cosx = 0\\
 \Rightarrow 2cosx\left( {2{{\cos }^2}x + 3\sqrt 2 \sin x - 4} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
\cos x = 0\\
2\left( {1 - {{\sin }^2}x} \right) + 3\sqrt 2 \sin x - 4 = 0
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
2{\sin ^2}x - 3\sqrt 2 \sin x + 2 = 0
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
\sin x = \sqrt 2 \left( {ktm} \right)\\
\sin x = \dfrac{{\sqrt 2 }}{2}
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = \dfrac{\pi }{4} + k2\pi \\
x = \dfrac{{3\pi }}{4} + k2\pi 
\end{array} \right.\\
4)\sin \dfrac{{4x}}{2} + \cos \dfrac{{4x}}{2} = 1 - 2\sin x\\
 \Rightarrow \sin 2x + \cos 2x = 1 - 2\sin x\\
 \Rightarrow 2\sin x.\cos x + 1 - 2{\sin ^2}x = 1 - 2{\mathop{\rm sinx}\nolimits} \\
 \Rightarrow 2sinx\left( {\cos x - \sin x + 1} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
\sin x = 0\\
\sin x - \cos x = 1
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
\sin x = 0\\
\sin \left( {x - \dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = k\pi \\
x - \dfrac{\pi }{4} = \dfrac{\pi }{4} + k2\pi \\
x - \dfrac{\pi }{4} = \dfrac{{3\pi }}{4} + k2\pi 
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{\pi }{2} + k2\pi \\
x = \pi  + k2\pi 
\end{array} \right. \Rightarrow x = \dfrac{{k\pi }}{2}
\end{array}$