Giải phương trình:a,cos2x -sin3x=0; b,tan3x×tan2x=-1 ; c,sin2x+sin6x=0; d,cot5x ×cot4x=1

\[\begin{array}{l} a)\\ \cos 2x - \sin 3x = 0 \Leftrightarrow \cos 2x = \sin 3x = \cos \left( {\frac{\pi }{2} - 3x} \right)\\ \Leftrightarrow \left[ \begin{array}{l} 2x = \frac{\pi }{2} - 3x + k2\pi \\ 2x = - \frac{\pi }{2} + 3x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \frac{\pi }{{10}} + \frac{{k2\pi }}{5}\\ x = \frac{\pi }{2} - k2\pi \end{array} \right.\\ b)\tan 3x.\tan 2x = - 1\left( {DK:\left\{ \begin{array}{l} \cos 3x \ne 0\\ \cos 2x \ne 0 \end{array} \right.} \right)\\ \Leftrightarrow \tan 3x = - \frac{1}{{\tan 2x}} = - \cot 2x = \cot \left( { - 2x} \right) = \tan \left( {\frac{\pi }{2} + 2x} \right)\\ \Leftrightarrow 3x = \frac{\pi }{2} + 2x + k\pi \Leftrightarrow x = \frac{\pi }{2} + k\pi \left( {TM} \right)\\ c)\sin 2x + \sin 6x = 0\\ \Leftrightarrow \sin 6x = - \sin 2x = \sin \left( { - 2x} \right)\\ \Leftrightarrow \left[ \begin{array}{l} 6x = - 2x + k2\pi \\ 6x = \pi + 2x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \frac{{k\pi }}{4}\\ x = \frac{\pi }{4} + \frac{{k\pi }}{2} \end{array} \right. \Leftrightarrow x = \frac{{k\pi }}{4}\\ d)\cot 5x.\cot 4x = 1\left( {DK:\left\{ \begin{array}{l} \sin 5x \ne 0\\ \sin 4x \ne 0 \end{array} \right.} \right)\\ \Leftrightarrow \cot 5x = \frac{1}{{\cot 4x}} = \tan 4x = \cot \left( {\frac{\pi }{2} - 4x} \right)\\ \Leftrightarrow 5x = \frac{\pi }{2} - 4x + k\pi \Leftrightarrow x = \frac{\pi }{{18}} + \frac{{k\pi }}{9}\left( {TM} \right) \end{array}\]