Đáp án:
$\begin{array}{l}
a)\cos 4x + \sin 5x = 0\\
\Rightarrow \cos 4x = - \sin 5x\\
\Rightarrow \cos 4x = \cos \left( {5x - \dfrac{\pi }{2}} \right)\\
\Rightarrow \left[ \begin{array}{l}
4x = 5x - \dfrac{\pi }{2} + k2\pi \\
4x = - 5x + \dfrac{\pi }{2} + k2\pi
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} - k2\pi \\
x = \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{9}
\end{array} \right.\\
b)\sin 2x + \cos 2x = \sqrt 2 \sin 3x\\
\Rightarrow \sqrt 2 .\sin \left( {2x + \dfrac{\pi }{4}} \right) = \sqrt 2 \sin 3x\\
\Rightarrow sin\left( {2x + \dfrac{\pi }{4}} \right) = \sin 3x\\
\Rightarrow \left[ \begin{array}{l}
2x + \dfrac{\pi }{4} = 3x + k2\pi \\
2x + \dfrac{\pi }{4} = \pi - 3x + k2\pi
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} - k2\pi \\
x = \dfrac{{3\pi }}{{20}} + \dfrac{{k2\pi }}{5}
\end{array} \right.
\end{array}$
