Giải phương trình: \(\tan \left( {{{120}^0} + 3x} \right) - \tan \left( {{{140}^0} - x} \right) = 2\sin \left( {{{80}^0} + 2x} \right)\)

ĐK:

\(\begin{array}{l}\left\{ \begin{array}{l}{120^0} + 3x \ne {90^0} + k{.180^0}\\{140^0} - x \ne {90^0} + k{.180^0}\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}3x \ne  - {30^0} + k{.180^0}\\x \ne {50^0} - k{.180^0}\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \ne  - {10^0} + k{.180^0}\\x \ne {50^0} - k{.180^0}\end{array} \right.\end{array}\)

\(\begin{array}{l}\tan \left( {{{120}^0} + 3x} \right) - \tan \left( {{{140}^0} - x} \right) = 2\sin \left( {{{80}^0} + 2x} \right)\\ \Leftrightarrow \tan \left[ {3\left( {{{40}^0} + x} \right)} \right] - \tan \left[ {{{180}^0} - \left( {{{40}^0} + x} \right)} \right] = 2\sin \left[ {2\left( {{{40}^0} + x} \right)} \right]\\ \Leftrightarrow \tan \left[ {3\left( {{{40}^0} + x} \right)} \right] - \tan \left[ { - \left( {{{40}^0} + x} \right)} \right] = 2\sin \left[ {2\left( {{{40}^0} + x} \right)} \right]\\ \Leftrightarrow \tan \left[ {3\left( {{{40}^0} + x} \right)} \right] + \tan \left( {{{40}^0} + x} \right) = 2\sin \left[ {2\left( {{{40}^0} + x} \right)} \right]\end{array}\)

Đặt \({40^0} + x = y\) ta được:

\(\begin{array}{l}\tan 3y + \tan y = 2\sin 2y\\ \Leftrightarrow \frac{{\sin 3y}}{{\cos 3y}} + \frac{{\sin y}}{{\cos y}} = 2\sin 2y\\ \Leftrightarrow \frac{{\sin 3y\cos y + \sin y\cos 3y}}{{\cos 3y\cos y}} = \frac{{2\sin 2y\cos 3y\cos y}}{{\cos 3y\cos y}}\\ \Rightarrow \sin 3y\cos y + \sin y\cos 3y = 2\sin 2y\cos 3y\cos y\\ \Leftrightarrow \sin 4y - 2\sin 2y\cos 3y\cos y = 0\\ \Leftrightarrow 2\sin 2y\cos 2y - 2\sin 2y\cos 3y\cos y = 0\\ \Leftrightarrow 2\sin 2y\left( {\cos 2y - \cos 3y\cos y} \right) = 0\\ \Leftrightarrow 2\sin 2y\left[ {\cos 2y - \frac{1}{2}\left( {\cos 4y + \cos 2y} \right)} \right] = 0\\ \Leftrightarrow 2\sin 2y\left( {\frac{1}{2}\cos 2y - \frac{1}{2}\cos 4y} \right) = 0\\ \Leftrightarrow \sin 2y\left( {\cos 2y - \cos 4y} \right) = 0\\ \Leftrightarrow \sin 2y.\left[ { - 2\sin 3y\sin \left( { - y} \right)} \right] = 0\\ \Leftrightarrow 2\sin y\sin 2y\sin 3y = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin y = 0\\\sin 2y = 0\\\sin 3y = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}y = k{180^0}\\2y = k{180^0}\\3y = k{180^0}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}y = k{.180^0}\\y = k{.90^0}\\y = k{.60^0}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}y = k{.60^0}\\y = {90^0} + k{.180^0}\end{array} \right.\end{array}\)

Suy ra

\(\begin{array}{l}\left[ \begin{array}{l}x + {40^0} = k{.60^0}\\x + {40^0} = {90^0} + k{.180^0}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x =  - {40^0} + k{.60^0}\\x = {50^0} + k{180^0}\left( {loai} \right)\end{array} \right.\\ \Leftrightarrow x =  - {40^0} + k{.60^0},k \in \mathbb{Z}\end{array}\)

Vậy pt có nghiệm \(x =  - {40^0} + k{.60^0},k \in \mathbb{Z}\).