Giải pt lượng giác sau a . tan^2 2x . tan^2 3x = 1 b. cot (3x + 2π/3 ). tan (x - π/3 ) = 1 Giúp giùm e vs

\[\begin{array}{l} a)\,\,{\tan ^2}2x.{\tan ^2}3x = 1\,\,\,\left( * \right)\\ DK:\,\,\left\{ \begin{array}{l} \cos 2x \ne 0\\ \cos 3x \ne 0 \end{array} \right..\\ \Rightarrow \left( * \right) \Leftrightarrow \frac{{{{\sin }^2}2x}}{{{{\cos }^2}2x}}.\frac{{{{\sin }^2}3x}}{{{{\cos }^2}3x}} = 1\\ \Leftrightarrow {\left( {\sin 2x.\sin 3x} \right)^2} = {\left( {\cos 2x.\cos 3x} \right)^2}\\ \Leftrightarrow {\left[ {\frac{1}{2}\left( {\cos x - \cos 5x} \right)} \right]^2} = {\left[ {\frac{1}{2}\left( {\cos 5x + \cos x} \right)} \right]^2}\\ \Leftrightarrow {\left( {\cos x - \cos 5x} \right)^2} = {\left( {\cos x + \cos 5x} \right)^2}\\ \Leftrightarrow - 2\cos x.\cos 5x = 2\cos x.\cos 5x\\ \Leftrightarrow \cos x.\cos 5x = 0\\ \Leftrightarrow \left[ \begin{array}{l} \cos x = 0\\ \cos 5x = 0 \end{array} \right..\\ b)\,\,\,\cot \left( {3x + \frac{{2\pi }}{3}} \right).tan\left( {x - \frac{\pi }{3}} \right) = 1\,\,\,\left( * \right)\\ DK:\,\,\,\left\{ \begin{array}{l} \cos \left( {x - \frac{\pi }{3}} \right) \ne 0\\ \sin \left( {3x\_\frac{{2\pi }}{3}} \right) \ne 0 \end{array} \right.\\ \left( * \right) \Leftrightarrow \frac{{\cos \left( {3x + \frac{{2\pi }}{3}} \right)}}{{\sin \left( {3x\_\frac{{2\pi }}{3}} \right)}}.\frac{{\sin \left( {x - \frac{\pi }{3}} \right)}}{{\cos \left( {x - \frac{\pi }{3}} \right)}} = 1\\ \Leftrightarrow \cos \left( {3x + \frac{{2\pi }}{3}} \right).\sin \left( {x - \frac{\pi }{3}} \right) = \sin \left( {3x - \frac{{2\pi }}{3}} \right).\cos \left( {x - \frac{\pi }{3}} \right)\\ \Leftrightarrow \sin \left( {x - \frac{\pi }{3} - 3x - \frac{{2\pi }}{3}} \right) + \sin \left( {x - \frac{\pi }{3} + 3x + \frac{{2\pi }}{3}} \right) = \sin \left( {3x - \frac{{2\pi }}{3} - x + \frac{\pi }{3}} \right) + \sin \left( {3x - \frac{{2\pi }}{3} + x - \frac{\pi }{3}} \right)\\ \Leftrightarrow \sin \left( { - 2x - \pi } \right) + \sin \left( {4x + \frac{\pi }{3}} \right) = \sin \left( {2x - \frac{\pi }{3}} \right) + \sin \left( {4x - \pi } \right)\\ \Leftrightarrow \sin \left( {2x + \pi } \right) + \sin \left( {4x + \frac{\pi }{3}} \right) = \sin \left( {2x - \frac{\pi }{3}} \right) - \sin \left( {\pi - 4x} \right)\\ \Leftrightarrow - \sin 2x + \sin 4x.\cos \frac{\pi }{3} + \cos 4x.\sin \frac{\pi }{3} = \sin 2x.cos\frac{\pi }{3} - \cos 2x.\sin \frac{\pi }{3} - \sin 4x\\ \Leftrightarrow \frac{1}{2}\sin 4x + \frac{{\sqrt 3 }}{2}\cos 4x - \sin 2x = \frac{1}{2}\sin 2x - \frac{{\sqrt 3 }}{2}\cos 2x - \sin 4x\\ \Leftrightarrow \frac{3}{2}\sin 4x + \frac{{\sqrt 3 }}{2}\cos 4x = \frac{3}{2}\sin 2x - \frac{{\sqrt 3 }}{2}\cos 2x\\ \Leftrightarrow \frac{{\sqrt 3 }}{2}\sin 4x + \frac{1}{2}\cos 4x = \frac{{\sqrt 3 }}{2}\sin 2x - \frac{1}{2}\cos 2x\\ \Leftrightarrow \sin \left( {4x + \frac{\pi }{6}} \right) = \sin \left( {2x - \frac{\pi }{6}} \right). \end{array}\] Em giải các phương trình lượng giác cơ bản nhé.