a) ta có : \(tanx=cot\left(\dfrac{\pi}{4}-2x\right)\Leftrightarrow tanx=cot\left(\dfrac{\pi}{2}-\left(\dfrac{\pi}{4}+2x\right)\right)\)

\(\Leftrightarrow tanx=tan\left(\dfrac{\pi}{4}+2x\right)\Leftrightarrow x=\dfrac{\pi}{4}+2x+k\pi\Leftrightarrow x=\dfrac{-\pi}{4}-k\pi\)

vậy phương trình này có một hệ nghiệm là : \(x=\dfrac{-\pi}{4}-k\pi\)

b) ta có : \(sin2x=cos3x\Leftrightarrow sin2x=cos\left(\dfrac{\pi}{2}-\left(\dfrac{\pi}{2}-3x\right)\right)\) \(\Leftrightarrow sin2x=sin\left(\dfrac{\pi}{2}-3x\right)\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{2}-3x+k2\pi\\2x=\pi-\dfrac{\pi}{2}+3x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{\pi}{2}+k2\pi\\-x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{10}+\dfrac{5k\pi}{2}\\x=\dfrac{-\pi}{2}-k2\pi\end{matrix}\right.\)

vậy phương trình này có 2 hệ nghiệm là : \(x=\dfrac{\pi}{10}+\dfrac{5k\pi}{2}vàx=\dfrac{-\pi}{2}-k2\pi\)

c) ta có : \(sin5x=-sin2x\Leftrightarrow sin5x=sin\left(-2x\right)\Leftrightarrow\left[{}\begin{matrix}5x=-2x+k2\pi\\5x=\pi+2x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}7x=k2\pi\\3x=\pi+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2k\pi}{7}\\x=\dfrac{\pi}{3}+\dfrac{2k\pi}{3}\end{matrix}\right.\)

vậy phương trình này có 2 hệ nghiệm : \(x=\dfrac{2k\pi}{7}vàx=\dfrac{\pi}{3}+\dfrac{2k\pi}{3}\)