Đáp án:

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

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
\cos \left( {\dfrac{\pi }{6} - x} \right) = \dfrac{1}{2}\\
 \Leftrightarrow \cos \left( {\dfrac{\pi }{6} - x} \right) = \cos \dfrac{\pi }{3}\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{\pi }{6} - x = \dfrac{\pi }{3} + k2\pi \\
\dfrac{\pi }{6} - x =  - \dfrac{\pi }{3} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{6} - \dfrac{\pi }{3} + k2\pi \\
x = \dfrac{\pi }{6} - \left( { - \dfrac{\pi }{3}} \right) + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{\pi }{2} + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
2,\\
\cos 2x + \sqrt 3 \sin 2x = 2\\
 \Leftrightarrow \dfrac{1}{2}\cos 2x + \dfrac{{\sqrt 3 }}{2}\sin 2x = 1\\
 \Leftrightarrow \cos 2x.\cos \dfrac{\pi }{3} + \sin 2x.\sin \dfrac{\pi }{3} = 1\\
 \Leftrightarrow \cos \left( {2x - \dfrac{\pi }{3}} \right) = 1\\
 \Leftrightarrow \cos \left( {2x - \dfrac{\pi }{2}} \right) = \cos 0\\
 \Leftrightarrow 2x - \dfrac{\pi }{2} = k2\pi \\
 \Leftrightarrow 2x = \dfrac{\pi }{2} + k2\pi \\
 \Leftrightarrow x = \dfrac{\pi }{4} + k\pi \,\,\,\,\left( {k \in Z} \right)\\
3,\\
2{\cos ^2}x + 3\cos x + 1 = 0\\
 \Leftrightarrow \left( {2{{\cos }^2}x + 2\cos x} \right) + \left( {\cos x + 1} \right) = 0\\
 \Leftrightarrow 2\cos x.\left( {\cos x + 1} \right) + \left( {\cos x + 1} \right) = 0\\
 \Leftrightarrow \left( {\cos x + 1} \right)\left( {2\cos x + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x + 1 = 0\\
2\cos x + 1 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x =  - 1\\
\cos x =  - \dfrac{1}{2}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x = \cos \pi \\
\cos x = \cos \dfrac{{2\pi }}{3}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \pi  + k2\pi \\
x = \dfrac{{2\pi }}{3} + k2\pi \\
x =  - \dfrac{{2\pi }}{3} + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)