Đáp án:

\(\begin{array}{l}
1,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k2\pi \\
x =  - \dfrac{\pi }{6} + k2\pi 
\end{array} \right.\,\,\,\,\,\,\,\,\left( {k \in Z} \right)\\
2,\\
x =  - \dfrac{\pi }{{18}} + \dfrac{{k\pi }}{3}\,\,\,\,\,\left( {k \in Z} \right)\\
3,\\
x = \dfrac{{5\pi }}{6} + k\pi \,\,\,\,\,\,\,\left( {k \in Z} \right)\\
4,\\
Phương\,\,trình\,\,vô\,\,nghiệm\\
5,\\
\left[ \begin{array}{l}
x =  - \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{7\pi }}{6} + k2\pi 
\end{array} \right.\,\,\,\,\,\,\left( {k \in Z} \right)\\
6,\\
\left[ \begin{array}{l}
x = k\pi \\
x = \arccos \dfrac{{ - 1}}{6} + k2\pi \\
x =  - \arccos \dfrac{{ - 1}}{6} + k2\pi 
\end{array} \right.\,\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\) 

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
2\cos x - \sqrt 3  = 0\\
 \Leftrightarrow 2\cos x = \sqrt 3 \\
 \Leftrightarrow \cos x = \dfrac{{\sqrt 3 }}{2}\\
 \Leftrightarrow \cos x = \cos \dfrac{\pi }{6}\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k2\pi \\
x =  - \dfrac{\pi }{6} + k2\pi 
\end{array} \right.\,\,\,\,\,\,\,\,\left( {k \in Z} \right)\\
2,\\
DKXD:\,\,\,\cos \left( {3x + \dfrac{\pi }{2}} \right) \ne 0 \Leftrightarrow 3x + \dfrac{\pi }{2} \ne \dfrac{\pi }{2} + k\pi  \Leftrightarrow 3x \ne k\pi  \Leftrightarrow x \ne \dfrac{{k\pi }}{3}\\
\sqrt 3 \tan \left( {3x + \dfrac{\pi }{2}} \right) - 3 = 0\\
 \Leftrightarrow \sqrt 3 \tan \left( {3x + \dfrac{\pi }{2}} \right) = 3\\
 \Leftrightarrow \tan \left( {3x + \dfrac{\pi }{2}} \right) = \sqrt 3 \\
 \Leftrightarrow \tan \left( {3x + \dfrac{\pi }{2}} \right) = \tan \dfrac{\pi }{3}\\
 \Leftrightarrow 3x + \dfrac{\pi }{2} = \dfrac{\pi }{3} + k\pi \\
 \Leftrightarrow 3x = \dfrac{\pi }{3} - \dfrac{\pi }{2} + k\pi \\
 \Leftrightarrow 3x =  - \dfrac{\pi }{6} + k\pi \\
 \Leftrightarrow x =  - \dfrac{\pi }{{18}} + \dfrac{{k\pi }}{3}\,\,\,\,\,\left( {k \in Z} \right)\\
3,\\
{\sin ^2}\left( {x - \dfrac{\pi }{3}} \right) + 2\cos \left( {x - \dfrac{\pi }{3}} \right) = 1\\
 \Leftrightarrow \left( {1 - {{\cos }^2}\left( {x - \dfrac{\pi }{3}} \right)} \right) + 2\cos \left( {x - \dfrac{\pi }{3}} \right) = 1\\
 \Leftrightarrow 1 - {\cos ^2}\left( {x - \dfrac{\pi }{3}} \right) + 2\cos \left( {x - \dfrac{\pi }{3}} \right) = 1\\
 \Leftrightarrow  - {\cos ^2}\left( {x - \dfrac{\pi }{3}} \right) + 2\cos \left( {x - \dfrac{\pi }{3}} \right) = 0\\
 \Leftrightarrow  - \cos \left( {x - \dfrac{\pi }{3}} \right).\left[ {\cos \left( {x - \dfrac{\pi }{3}} \right) - 2} \right] = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos \left( {x - \dfrac{\pi }{3}} \right) = 0\\
\cos \left( {x - \dfrac{\pi }{3}} \right) - 2 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos \left( {x - \dfrac{\pi }{3}} \right) = 0\\
\cos \left( {x - \dfrac{\pi }{3}} \right) = 2
\end{array} \right.\\
 - 1 \le \cos \left( {x - \dfrac{\pi }{3}} \right) \le 1\\
 \Rightarrow \cos \left( {x - \dfrac{\pi }{3}} \right) = 0\\
 \Leftrightarrow x - \dfrac{\pi }{3} = \dfrac{\pi }{2} + k\pi \\
 \Leftrightarrow x = \dfrac{\pi }{2} + \dfrac{\pi }{3} + k\pi \\
 \Leftrightarrow x = \dfrac{{5\pi }}{6} + k\pi \,\,\,\,\,\,\,\left( {k \in Z} \right)\\
4,\\
\cos 2x + 2\sin x - 3 = 0\\
 \Leftrightarrow \left( {1 - 2{{\sin }^2}x} \right) + 2\sin x - 3 = 0\\
 \Leftrightarrow 1 - 2{\sin ^2}x + 2\sin x - 3 = 0\\
 \Leftrightarrow  - 2{\sin ^2}x + 2\sin x - 2 = 0\\
 \Leftrightarrow {\sin ^2}x - \sin  + 1 = 0\\
 \Leftrightarrow \left( {{{\sin }^2}x - \sin x + \dfrac{1}{4}} \right) + \dfrac{3}{4} = 0\\
 \Leftrightarrow \left( {{{\sin }^2}x - 2.\sin .\dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \right) + \dfrac{3}{4} = 0\\
 \Leftrightarrow {\left( {\sin x - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} = 0\\
{\left( {\sin x - \dfrac{1}{2}} \right)^2} \ge 0,\,\,\,\forall \,x\\
 \Rightarrow {\left( {\sin x - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4} > 0,\,\,\,\,\forall \,x\\
 \Rightarrow Phương\,\,trình\,\,vô\,\,nghiệm\\
5,\\
6{\cos ^2}x + 5\sin x - 2 = 0\\
 \Leftrightarrow 6.\left( {1 - {{\sin }^2}x} \right) + 5\sin x - 2 = 0\\
 \Leftrightarrow 6 - 6{\sin ^2}x + 5\sin x - 2 = 0\\
 \Leftrightarrow  - 6{\sin ^2}x + 5\sin x + 4 = 0\\
 \Leftrightarrow 6{\sin ^2}x - 5\sin x - 4 = 0\\
 \Leftrightarrow \left( {6{{\sin }^2}x + 3\sin x} \right) + \left( { - 8\sin x - 4} \right) = 0\\
 \Leftrightarrow 3\sin x.\left( {2\sin x + 1} \right) - 4.\left( {2\sin x + 1} \right) = 0\\
 \Leftrightarrow \left( {2\sin x + 1} \right)\left( {3\sin x - 4} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
2\sin x + 1 = 0\\
3\sin x - 4 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x =  - \dfrac{1}{2}\\
\sin x = \dfrac{4}{3}
\end{array} \right.\\
 - 1 \le \sin x \le 1 \Rightarrow \sin x =  - \dfrac{1}{2}\\
\sin x =  - \dfrac{1}{2}\\
 \Leftrightarrow \sin x = \sin \left( { - \dfrac{\pi }{6}} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{\pi }{6} + k2\pi \\
x = \pi  - \left( {\dfrac{\pi }{6}} \right) + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{7\pi }}{6} + k2\pi 
\end{array} \right.\,\,\,\,\,\,\left( {k \in Z} \right)\\
6,\\
\sin x + 3\sin 2x = 0\\
 \Leftrightarrow \sin x + 3.\left( {2\sin x.\cos x} \right) = 0\\
 \Leftrightarrow \sin x + 6\sin x.\cos x = 0\\
 \Leftrightarrow \sin x\left( {1 + 6\cos x} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
1 + 6\cos x = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\cos x =  - \dfrac{1}{6}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \arccos \dfrac{{ - 1}}{6} + k2\pi \\
x =  - \arccos \dfrac{{ - 1}}{6} + k2\pi 
\end{array} \right.\,\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)