Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
DKXD:\left\{ \begin{array}{l}
\cos x \ne 0\\
\cos 2x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne \dfrac{\pi }{2} + k\pi \\
2x \ne \dfrac{\pi }{2} + k\pi 
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne \dfrac{\pi }{2} + k\pi \\
x \ne \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}
\end{array} \right.\\
\tan x + \tan 2x = \sin 3x.\cos x\\
 \Leftrightarrow \dfrac{{\sin x}}{{\cos x}} + \dfrac{{\sin 2x}}{{\cos 2x}} = \sin 3x.\cos x\\
 \Leftrightarrow \dfrac{{\sin x.\cos 2x + \sin 2x.\cos x}}{{\cos x.\cos 2x}} = \sin 3x.\cos x\\
 \Leftrightarrow \dfrac{{\sin \left( {x + 2x} \right)}}{{\cos x.\cos 2x}} = \sin 3x.\cos x\\
 \Leftrightarrow \dfrac{{\sin 3x}}{{\cos x.\cos 2x}} = \sin 3x.\cos x\\
 \Leftrightarrow \sin 3x.\left( {\dfrac{1}{{\cos x.\cos 2x}} - \cos x} \right) = 0\\
 \Leftrightarrow \sin 3x.\dfrac{{1 - {{\cos }^2}x.\cos 2x}}{{\cos x.\cos 2x}} = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin 3x = 0\\
1 - {\cos ^2}x.\cos 2x = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
3x = k\pi \\
1 - {\cos ^2}x.\left( {2{{\cos }^2}x - 1} \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{3}\\
 - 2{\cos ^4}x + {\cos ^2}x + 1 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{3}\\
\left( {{{\cos }^2}x - 1} \right)\left( {2{{\cos }^2}x + 1} \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{3}\\
{\cos ^2}x = 1\\
{\cos ^2}x = \dfrac{{ - 1}}{2}\,\,\,\,\left( L \right)
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{3}\\
{\cos ^2}x = 1
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{3}\\
{\sin ^2}x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{3}\\
x = k\pi 
\end{array} \right. \Leftrightarrow x = \dfrac{{k\pi }}{3}\,\,\,\,\,\,\left( {k \in Z} \right)\\
b,\\
{\cos ^2}x - {\sin ^2}x = \sin 3x + \cos 4x\\
 \Leftrightarrow \cos 2x = \sin 3x + \cos 4x\\
 \Leftrightarrow \cos 2x - \cos 4x = \sin 3x\\
 \Leftrightarrow  - 2.\sin \dfrac{{2x + 4x}}{2}.\sin \dfrac{{2x - 4x}}{2} = \sin 3x\\
 \Leftrightarrow  - 2.\sin 3x.\sin \left( { - x} \right) = \sin 3x\\
 \Leftrightarrow 2\sin 3x.\sin x = \sin 3x\\
 \Leftrightarrow \sin 3x.\left( {2\sin x - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin 3x = 0\\
\sin x = \dfrac{1}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
3x = k\pi \\
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{3}\\
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)