giải pt: sin3x+cos5x=0

Đáp án:

$\begin{array}{l}
\sin 3x + \cos 5x = 0\\
 \Leftrightarrow \sin 3x =  - \cos 5x\\
 \Leftrightarrow \sin 3x = \cos \left( {5x + \pi } \right)\\
 \Leftrightarrow \sin 3x = \sin \left( {\dfrac{\pi }{2} - 5x - \pi } \right)\\
 \Leftrightarrow \sin 3x = \sin \left( { - \dfrac{\pi }{2} - 5x} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
3x =  - \dfrac{\pi }{2} - 5x + k2\pi \\
3x = \pi  + \dfrac{\pi }{2} + 5x + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{\pi }{{16}} + \dfrac{{k\pi }}{4}\\
x =  - \dfrac{{3\pi }}{4} - k\pi 
\end{array} \right.\\
Vậy\,\left[ \begin{array}{l}
x =  - \dfrac{\pi }{{16}} + \dfrac{{k\pi }}{4}\\
x =  - \dfrac{{3\pi }}{4} - k\pi 
\end{array} \right.
\end{array}$