1 câu trả lời
\[\begin{array}{l} \tan 3x.\tan 2x = 1\\ DK:\,\,\,\,\left\{ \begin{array}{l} \cos 3x \ne 0\\ \cos 2x \ne 0 \end{array} \right.\\ \Leftrightarrow \frac{{\sin 3x}}{{\cos 3x}}.\frac{{\sin 2x}}{{\cos 2x}} = 1\\ \Leftrightarrow \sin 3x.\sin 2x = \cos 3x.\cos 2x\\ \Leftrightarrow \frac{1}{2}\left( {\cos x - \cos 5x} \right) = \frac{1}{2}\left( {\cos x + \cos 5x} \right)\\ \Leftrightarrow \cos x - \cos 5x = \cos x + \cos 5x\\ \Leftrightarrow 2\cos 5x = 0\\ \Leftrightarrow \cos 5x = 0\\ \Leftrightarrow 5x = \frac{\pi }{2} + k\pi \\ \Leftrightarrow x = \frac{\pi }{{10}} + \frac{{k\pi }}{5}\,\,\,\left( {k \in Z} \right). \end{array}\]
