Đáp án:

\(\begin{array}{l}
1,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\,\,\,\left( {k \in Z} \right)\\
2,\\
Phương\,\,trình\,\,vô\,\,nghiệm\\
3,\\
x =  - \dfrac{\pi }{6} + k\pi \,\,\,\left( {k \in Z} \right)\\
4,\\
x = \dfrac{\pi }{6} + k\pi \,\,\,\left( {k \in Z} \right)\\
5,\\
\left[ \begin{array}{l}
x =  - \dfrac{\pi }{6} + k\pi \\
x = \dfrac{{2\pi }}{3} + k\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
6,\\
\left[ \begin{array}{l}
x = \dfrac{{5\pi }}{{12}} + \dfrac{{k2\pi }}{3}\\
x =  - \dfrac{\pi }{{12}} + \dfrac{{k2\pi }}{3}\,
\end{array} \right.\,\,\,\left( {k \in Z} \right)
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
2\sin x - 1 = 0\\
 \Leftrightarrow \sin x = \dfrac{1}{2}\\
 \Leftrightarrow \sin x = \sin \dfrac{\pi }{6}\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k2\pi \\
x = \pi  - \dfrac{\pi }{6} + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\,\,\,\left( {k \in Z} \right)\\
2,\\
\sqrt 2 \cos x + 2 = 0\\
 \Leftrightarrow \cos x =  - \sqrt 2 \\
 - 1 \le \cos x \le 1 \Rightarrow Phương\,\,trình\,\,vô\,\,nghiệm\\
3,\\
DKXD:\,\,\cos x \ne 0 \Leftrightarrow x \ne \dfrac{\pi }{2} + k\pi \,\,\,\left( {k \in Z} \right)\\
\sqrt 3 \tan x + 1 = 0\\
 \Leftrightarrow \tan x =  - \dfrac{1}{{\sqrt 3 }}\\
 \Leftrightarrow \tan x = \tan \dfrac{{ - \pi }}{6}\\
 \Leftrightarrow x =  - \dfrac{\pi }{6} + k\pi \,\,\,\left( {k \in Z} \right)\\
4,\\
DKXD:\,\,\,\sin x \ne 0 \Leftrightarrow x \ne k\pi \,\,\,\left( {k \in Z} \right)\\
\cot x - \sqrt 3  = 0\\
 \Leftrightarrow \cot x = \sqrt 3 \\
 \Leftrightarrow \cot x = \cot \dfrac{\pi }{6}\\
 \Leftrightarrow x = \dfrac{\pi }{6} + k\pi \,\,\,\left( {k \in Z} \right)\\
5,\\
2\sin 2x + \sqrt 3  = 0\\
 \Leftrightarrow \sin 2x =  - \dfrac{{\sqrt 3 }}{2}\\
 \Leftrightarrow \sin 2x = \sin \dfrac{{ - \pi }}{3}\\
 \Leftrightarrow \left[ \begin{array}{l}
2x =  - \dfrac{\pi }{3} + k2\pi \\
2x = \pi  - \left( { - \dfrac{\pi }{3}} \right) + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x =  - \dfrac{\pi }{3} + k2\pi \\
2x = \dfrac{{4\pi }}{3} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{\pi }{6} + k\pi \\
x = \dfrac{{2\pi }}{3} + k\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
6,\\
2\cos \left( {3x - \dfrac{\pi }{2}} \right) + \sqrt 2  = 0\\
 \Leftrightarrow \cos \left( {3x - \dfrac{\pi }{2}} \right) =  - \dfrac{{\sqrt 2 }}{2}\\
 \Leftrightarrow \cos \left( {3x - \dfrac{\pi }{2}} \right) = \cos \dfrac{{3\pi }}{4}\\
 \Leftrightarrow \left[ \begin{array}{l}
3x - \dfrac{\pi }{2} = \dfrac{{3\pi }}{4} + k2\pi \\
3x - \dfrac{\pi }{2} =  - \dfrac{{3\pi }}{4} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
3x = \dfrac{{5\pi }}{4} + k2\pi \\
3x =  - \dfrac{\pi }{4} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{5\pi }}{{12}} + \dfrac{{k2\pi }}{3}\\
x =  - \dfrac{\pi }{{12}} + \dfrac{{k2\pi }}{3}\,
\end{array} \right.\,\,\,\left( {k \in Z} \right)
\end{array}\)