Đáp án:

$\begin{array}{l}
a)Khi:\sin 2x = 0\\
 \Rightarrow 2x = k\pi \\
 \Rightarrow x = \dfrac{{k\pi }}{2}\\
Pt:2{\cos ^2}2x =  - 4\\
 \Rightarrow {\cos ^2}2x =  - 4\left( {ktm} \right)\\
 + Khi:\sin 2x \ne 0 \Rightarrow x \ne \dfrac{{k\pi }}{2}\\
Pt:2{\cos ^2}2x - 3\sqrt 3 .2\sin 2x.\cos 2x - 4{\sin ^2}2x =  - 4\\
 \Rightarrow \dfrac{{2{{\cos }^2}2x}}{{{{\sin }^2}2x}} - \dfrac{{3\sqrt 3 \cos 2x}}{{\sin 2x}} - 4 = \dfrac{{ - 4}}{{{{\sin }^2}2x}}\\
 \Rightarrow 2{\cot ^2}2x - 3\sqrt 3 \cot 2x - 4 =  - 4.\dfrac{1}{{{{\sin }^2}2x}}\\
 \Rightarrow 2{\cot ^2}2x - 3\sqrt 3 \cot 2x - 4 + 4.\left( {{{\cot }^2}2x + 1} \right) = 0\\
 \Rightarrow 6{\cot ^2}2x - 3\sqrt 3 \cot 2x = 0\\
 \Rightarrow 3\cot 2x\left( {2\cot 2x - \sqrt 3 } \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
\cot 2x = 0\\
\cot 2x = \dfrac{{\sqrt 3 }}{2}
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
2x = {90^0} + k{.180^0}\\
2x = {49^0} + k{.180^0}
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = {45^0} + k{.90^0}\\
x = 24,{5^0} + k{.90^0}
\end{array} \right.\left( {tmdk} \right)\\
b){\sin ^2}x + \sin 2x - 2{\cos ^2}x = \dfrac{1}{2}\\
 + Khi:\cos x = 0 \Rightarrow x = \dfrac{\pi }{2} + k\pi \\
 \Rightarrow {\sin ^2}2x = \dfrac{1}{2}\left( {ktm} \right)\\
 + Khi:\cos x \ne 0 \Rightarrow x \ne \dfrac{\pi }{2} + k\pi \\
 \Rightarrow \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + \dfrac{{2\sin x.\cos x}}{{{{\cos }^2}x}} - 2 = \dfrac{1}{2}.\dfrac{1}{{{{\cos }^2}x}}\\
 \Rightarrow {\tan ^2}x + 2\tan x - 2 = \dfrac{1}{2}\left( {{{\tan }^2}x + 1} \right)\\
 \Rightarrow \dfrac{1}{2}{\tan ^2}x + 2\tan x - \dfrac{5}{2} = 0\\
 \Rightarrow {\tan ^2}x + 4\tan x - 5 = 0\\
 \Rightarrow \left( {\tan x - 1} \right)\left( {\tan x + 5} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
\tan x = 1\\
\tan x =  - 5
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \left( {tm} \right)\\
x = \arctan \left( { - 5} \right) + k\pi \left( {tm} \right)
\end{array} \right.
\end{array}$