Đáp án:

\(\begin{array}{l}
1,\\
\left[ \begin{array}{l}
x = 60^\circ  + k.360^\circ \\
x = k.180^\circ \\
x =  - 120^\circ  + k.360^\circ 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
2,\\
x = \dfrac{\pi }{6} + \dfrac{{k\pi }}{2}\,\,\,\,\,\left( {k \in Z} \right)\\
3,\\
x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
1,\\
{\cos ^2}\left( {x - 30^\circ } \right) = \dfrac{3}{4}\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos \left( {x - 30^\circ } \right) = \dfrac{{\sqrt 3 }}{2}\\
\cos \left( {x - 30^\circ } \right) =  - \dfrac{{\sqrt 3 }}{2}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 30^\circ  = 30^\circ  + k.360^\circ \\
x - 30^\circ  =  - 30^\circ  + k.360^\circ \\
x - 30^\circ  = 150^\circ  + k.360^\circ \\
x - 30^\circ  =  - 150^\circ  + k.360^\circ 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 60^\circ  + k.360^\circ \\
x = k.360^\circ \\
x = 180^\circ  + k.360^\circ \\
x =  - 120^\circ  + k.360^\circ 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 60^\circ  + k.360^\circ \\
x = k.180^\circ \\
x =  - 120^\circ  + k.360^\circ 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
2,\\
\sin \left( {2x + \dfrac{\pi }{3}} \right) = 2\sin x.\cos x\\
 \Leftrightarrow \sin \left( {2x + \dfrac{\pi }{3}} \right) = \sin 2x\\
 \Leftrightarrow \left[ \begin{array}{l}
2x + \dfrac{\pi }{3} = 2x + k2\pi \\
2x + \dfrac{\pi }{3} = \pi  - 2x + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
0x =  - \dfrac{\pi }{3} + k2\pi \,\,\,\,\left( L \right)\\
4x = \dfrac{{2\pi }}{3} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow x = \dfrac{\pi }{6} + \dfrac{{k\pi }}{2}\\
3,\\
2{\cos ^2}x - 1 = 2\sin x.\cos x\\
 \Leftrightarrow \cos 2x = \sin 2x\\
 \Leftrightarrow \cos 2x = \cos \left( {\dfrac{\pi }{2} - 2x} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} - 2x + k2\pi \\
2x = 2x - \dfrac{\pi }{2} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
4x = \dfrac{\pi }{2} + k2\pi \\
0x =  - \dfrac{\pi }{2} + k2\pi \,\,\,\,\,\,\left( L \right)
\end{array} \right.\\
 \Leftrightarrow x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)