Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
\sqrt 3 \sin 3x - \cos 3x = \sqrt 2 \\
 \Leftrightarrow \dfrac{{\sqrt 3 }}{2}.\sin 3x - \dfrac{1}{2}\cos 3x = \dfrac{{\sqrt 2 }}{2}\\
 \Leftrightarrow \sin 3x.\cos \dfrac{\pi }{6} - \cos 3x.\sin \dfrac{\pi }{6} = \dfrac{{\sqrt 2 }}{2}\\
 \Leftrightarrow \sin \left( {3x - \dfrac{\pi }{6}} \right) = \dfrac{{\sqrt 2 }}{2}\\
 \Leftrightarrow \left[ \begin{array}{l}
3x - \dfrac{\pi }{6} = \dfrac{\pi }{4} + k2\pi \\
3x - \dfrac{\pi }{6} = \dfrac{{3\pi }}{4} + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{5\pi }}{{36}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{11\pi }}{{36}} + \dfrac{{k2\pi }}{3}
\end{array} \right.\\
b,\\
\cos 7x - \sqrt 3 \sin 7x =  - \sqrt 2 \\
 \Leftrightarrow \dfrac{1}{2}.\cos 7x - \dfrac{{\sqrt 3 }}{2}\sin 7x =  - \dfrac{{\sqrt 2 }}{2}\\
 \Leftrightarrow \cos 7x.\cos \dfrac{\pi }{3} - \sin 7x.\sin \dfrac{\pi }{3} =  - \dfrac{{\sqrt 2 }}{2}\\
 \Leftrightarrow \cos \left( {7x + \dfrac{\pi }{3}} \right) =  - \dfrac{{\sqrt 2 }}{2}\\
 \Leftrightarrow \left[ \begin{array}{l}
7x + \dfrac{\pi }{3} = \dfrac{{3\pi }}{4} + k2\pi \\
7x + \dfrac{\pi }{3} =  - \dfrac{{3\pi }}{4} + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{5\pi }}{{84}} + \dfrac{{k2\pi }}{7}\\
x = \dfrac{{ - 13\pi }}{{84}} + \dfrac{{k2\pi }}{7}
\end{array} \right.\\
c,\\
\sqrt 3 {\sin ^2}x + \dfrac{1}{2}\sin 2x = \sqrt 3 \\
 \Leftrightarrow \sqrt 3 \left( {{{\sin }^2}x - 1} \right) + \dfrac{1}{2}.2\sin x.\cos x = 0\\
 \Leftrightarrow \sqrt 3 .\left( { - \cos x} \right) + \sin x.\cos x = 0\\
 \Leftrightarrow \cos x.\left( {\sin x - \sqrt 3 } \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\sin x = \sqrt 3 \,\,\,\,\left( {L,\,\,\, - 1 \le \sin x \le 1} \right)
\end{array} \right.\\
 \Leftrightarrow \cos x = 0\\
 \Leftrightarrow x = \dfrac{\pi }{2} + k\pi \\
d,\\
3{\cos ^2}x - 2\cos 2x = 3\sin x - 1\\
 \Leftrightarrow 3.\left( {1 - {{\sin }^2}x} \right) - 2.\left( {1 - 2{{\sin }^2}x} \right) = 3\sin x - 1\\
 \Leftrightarrow 3 - 3{\sin ^2}x - 2 + 4{\sin ^2}x - 3\sin x + 1 = 0\\
 \Leftrightarrow {\sin ^2}x - 3\sin x + 2 = 0\\
 \Leftrightarrow \left( {\sin x - 1} \right)\left( {\sin x - 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x = 1\\
\sin x = 2\,\,\,\,\left( {L;\,\,\, - 1 \le \sin x \le 1} \right)
\end{array} \right.\\
 \Leftrightarrow \sin x = 1\\
 \Leftrightarrow x = \dfrac{\pi }{2} + k2\pi \\
e,\\
2\cos 2x + 3\sin x - 1 = 0\\
 \Leftrightarrow 2.\left( {1 - 2{{\sin }^2}x} \right) + 3\sin x - 1 = 0\\
 \Leftrightarrow 2 - 4{\sin ^2}x + 3\sin x - 1 = 0\\
 \Leftrightarrow  - 4{\sin ^2}x + 3\sin x + 1 = 0\\
 \Leftrightarrow 4{\sin ^2}x - 3\sin x - 1 = 0\\
 \Leftrightarrow \left( {\sin x - 1} \right)\left( {4\sin x + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x = 1\\
\sin x =  - \dfrac{1}{4}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k2\pi \\
x = \arcsin \left( { - \dfrac{1}{4}} \right) + k2\pi \\
x = \pi  - \arcsin \left( { - \dfrac{1}{4}} \right) + k2\pi 
\end{array} \right.
\end{array}\)