Đáp án:

\(\begin{array}{l}
a,\\
\left[ \begin{array}{l}
x = \dfrac{{k\pi }}{2}\\
x = \dfrac{{3\pi }}{8} + k\pi \\
x =  - \dfrac{{3\pi }}{8} + k\pi 
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
b,\\
x = k4\pi \,\,\,\,\left( {k \in Z} \right)\\
c,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = \arctan \left( { - 2} \right) + k\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
2\sin 2x + \sqrt 2 \sin 4x = 0\\
 \Leftrightarrow 2\sin 2x + \sqrt 2 .2\sin 2x.\cos 2x = 0\\
 \Leftrightarrow 2\sin 2x.\left( {1 + \sqrt 2 \cos 2x} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin 2x = 0\\
1 + \sqrt 2 \cos 2x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sin 2x = 0\\
\cos 2x =  - \dfrac{1}{{\sqrt 2 }}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin 2x = 0\\
\cos 2x = \cos \dfrac{{3\pi }}{4}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = k\pi \\
2x = \dfrac{{3\pi }}{4} + k2\pi \\
2x =  - \dfrac{{3\pi }}{4} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{2}\\
x = \dfrac{{3\pi }}{8} + k\pi \\
x =  - \dfrac{{3\pi }}{8} + k\pi 
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
b,\\
{\sin ^2}\dfrac{x}{2} - 2\cos \dfrac{x}{2} + 2 = 0\\
 \Leftrightarrow \left( {1 - {{\cos }^2}\dfrac{x}{2}} \right) - 2\cos \dfrac{x}{2} + 2 = 0\\
 \Leftrightarrow  - {\cos ^2}\dfrac{x}{2} - 2\cos \dfrac{x}{2} + 3 = 0\\
 \Leftrightarrow {\cos ^2}\dfrac{x}{2} + 2\cos \dfrac{x}{2} - 3 = 0\\
 \Leftrightarrow \left( {\cos \dfrac{x}{2} - 1} \right)\left( {\cos \dfrac{x}{2} + 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos \dfrac{x}{2} - 1 = 0\\
\cos \dfrac{x}{2} + 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos \dfrac{x}{2} = 1\\
\cos \dfrac{x}{2} =  - 3
\end{array} \right.\\
 - 1 \le \cos \dfrac{x}{2} \le 1 \Rightarrow \cos \dfrac{x}{2} = 1\\
\cos \dfrac{x}{2} = 1\\
 \Leftrightarrow \dfrac{x}{2} = k2\pi \\
 \Leftrightarrow x = k4\pi \,\,\,\,\left( {k \in Z} \right)\\
c,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
\sin x \ne 0\\
\cos x \ne 0
\end{array} \right. \Leftrightarrow x \ne \dfrac{{k\pi }}{2}\,\,\,\left( {k \in Z} \right)\\
\tan x - 2\cot x + 1 = 0\\
 \Leftrightarrow \tan x - \dfrac{2}{{\tan x}} + 1 = 0\\
 \Leftrightarrow \dfrac{{{{\tan }^2}x - 2 + \tan x}}{{\tan x}} = 0\\
 \Leftrightarrow {\tan ^2}x + \tan x - 2 = 0\\
 \Leftrightarrow \left( {\tan x - 1} \right)\left( {\tan x + 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\tan x - 1 = 0\\
\tan x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\tan x = 1\\
\tan x =  - 2
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = \arctan \left( { - 2} \right) + k\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)