Đáp án:
\[\left[ \begin{array}{l}
x = \dfrac{\pi }{7} + \dfrac{{k2\pi }}{7}\\
x = - \dfrac{\pi }{3} + \dfrac{{k2\pi }}{3}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\cos x = - \cos \left( {\pi - x} \right)\\
\cos x = \cos y \Leftrightarrow \left[ \begin{array}{l}
x = y + k2\pi \\
x = - y + k2\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
\cos 5x + \cos 2x = 0\\
\Leftrightarrow \cos 5x = - \cos 2x\\
\Leftrightarrow \cos 5x = \cos \left( {\pi - 2x} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
5x = \pi - 2x + k2\pi \\
5x = 2x - \pi + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
5x + 2x = \pi + k2\pi \\
5x - 2x = - \pi + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
7x = \pi + k2\pi \\
3x = - \pi + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{7} + \dfrac{{k2\pi }}{7}\\
x = - \dfrac{\pi }{3} + \dfrac{{k2\pi }}{3}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
