Đáp án:
\[\left[ \begin{array}{l}
x = - \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\\
x = - \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{3}
\end{array} \right.\,\,\,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\sin 5x + \cos x = 0\\
\Leftrightarrow \cos x = - \sin 5x\\
\Leftrightarrow \cos x = \sin \left( { - 5x} \right)\\
\Leftrightarrow \cos x = \cos \left( {\dfrac{\pi }{2} - \left( { - 5x} \right)} \right)\\
\Leftrightarrow \cos x = \cos \left( {\dfrac{\pi }{2} + 5x} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + 5x + k2\pi \\
x = - \dfrac{\pi }{2} - 5x + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
- 4x = \dfrac{\pi }{2} + k2\pi \\
6x = - \dfrac{\pi }{2} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\\
x = - \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{3}
\end{array} \right.\,\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
