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

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