Đáp án:

\(\begin{array}{l}
a,\\
\left[ \begin{array}{l}
x =  - \dfrac{\pi }{{36}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{5\pi }}{{36}} + \dfrac{{k2\pi }}{3}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
b,\\
x =  \pm \dfrac{{2\pi }}{3} + k2\pi \,\,\,\,\left( {k \in Z} \right)\\
c,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k2\pi \\
x = \dfrac{\pi }{6} + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
\sin \left( {3x + \dfrac{\pi }{3}} \right) = \dfrac{{\sqrt 2 }}{2}\\
 \Leftrightarrow \sin \left( {3x + \dfrac{\pi }{3}} \right) = \sin \dfrac{\pi }{4}\\
 \Leftrightarrow \left[ \begin{array}{l}
3x + \dfrac{\pi }{3} = \dfrac{\pi }{4} + k2\pi \\
3x + \dfrac{\pi }{3} = \pi  - \dfrac{\pi }{4} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
3x + \dfrac{\pi }{3} = \dfrac{\pi }{4} + k2\pi \\
3x + \dfrac{\pi }{3} = \dfrac{{3\pi }}{4} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
3x =  - \dfrac{\pi }{{12}} + k2\pi \\
3x = \dfrac{{5\pi }}{{12}} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{\pi }{{36}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{5\pi }}{{36}} + \dfrac{{k2\pi }}{3}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
b,\\
2{\cos ^2}x - 7\cos x - 4 = 0\\
 \Leftrightarrow \left( {2{{\cos }^2}x - 8\cos x} \right) + \left( {\cos x - 4} \right) = 0\\
 \Leftrightarrow 2\cos x.\left( {\cos x - 4} \right) + \left( {\cos x - 4} \right) = 0\\
 \Leftrightarrow \left( {\cos x - 4} \right)\left( {2\cos x + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x - 4 = 0\\
2\cos x + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x = 4\\
\cos x =  - \dfrac{1}{2}
\end{array} \right.\\
 - 1 \le \cos x \le 1 \Leftrightarrow \cos x =  - \dfrac{1}{2}\\
\cos x =  - \dfrac{1}{2}\\
 \Leftrightarrow \cos x = \cos \dfrac{{2\pi }}{3}\\
 \Leftrightarrow x =  \pm \dfrac{{2\pi }}{3} + k2\pi \,\,\,\,\left( {k \in Z} \right)\\
c,\\
3\sin x + \sqrt 3 \cos x = 3\\
 \Leftrightarrow \dfrac{{\sqrt 3 }}{{2\sqrt 3 }}\cos x + \dfrac{3}{{2\sqrt 3 }}\sin x = \dfrac{3}{{2\sqrt 3 }}\\
 \Leftrightarrow \dfrac{1}{2}\cos x + \dfrac{{\sqrt 3 }}{2}\sin x = \dfrac{{\sqrt 3 }}{2}\\
 \Leftrightarrow \cos x.\cos \dfrac{\pi }{3} + \sin x.\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}\\
 \Leftrightarrow \cos \left( {x - \dfrac{\pi }{3}} \right) = \cos \dfrac{\pi }{6}\\
 \Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{\pi }{3} = \dfrac{\pi }{6} + k2\pi \\
x - \dfrac{\pi }{3} =  - \dfrac{\pi }{6} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k2\pi \\
x = \dfrac{\pi }{6} + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)