Đáp án:
$\begin{array}{l}
a)\sin 4x + \cos 5x = 0\\
\Rightarrow \sin 4x = - \cos 5x\\
\Rightarrow \sin 4x = cos\left( {5x + \pi } \right)\\
\Rightarrow \sin 4x = \sin \left( {\dfrac{\pi }{2} - 5x - \pi } \right)\\
\Rightarrow \left[ \begin{array}{l}
4x = - 5x - \dfrac{\pi }{2} + k2\pi \\
4x = \pi + 5x + \dfrac{\pi }{2} + k2\pi
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = - \dfrac{\pi }{{18}} + \dfrac{{k2\pi }}{9}\\
x = \dfrac{{ - 3\pi }}{2} - k2\pi
\end{array} \right.\\
\Rightarrow x = - \dfrac{\pi }{{18}}\\
b){\sin ^2}2x + {\cos ^2}5x = 1\\
\Rightarrow {\sin ^2}2x + {\cos ^2}5x = {\sin ^2}5x + {\cos ^2}5x\\
\Rightarrow {\sin ^2}2x = {\sin ^2}5x\\
\Rightarrow \left[ \begin{array}{l}
\sin 2x = sin5x\\
sin2x = - sin5x
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
2x = 5x + k2\pi \\
2x = \pi - 5x + k2\pi \\
2x = 5x + \pi + k2\pi \\
2x = - 5x + k2\pi
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{{k2\pi }}{3}\\
x = \dfrac{\pi }{7} + \dfrac{{k2\pi }}{7}\\
x = \dfrac{{ - \pi }}{3} - \dfrac{{k2\pi }}{3}\\
x = \dfrac{{k2\pi }}{7}
\end{array} \right.\\
\Rightarrow x = - \dfrac{\pi }{7}\left( {khi:k = - 1} \right)
\end{array}$
