Đáp án:
\(\left[ \begin{array}{l}
x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{4}\\
x = \dfrac{\pi }{{10}} + \dfrac{{k\pi }}{5}\\
x = \dfrac{\pi }{2} + k\pi
\end{array} \right.\,\,\left( {k \in Z} \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
\cos 2x + \cos 8x + \cos 10x = 1\\
\Leftrightarrow \left( {\cos 2x + \cos 10x} \right) + \cos 8x - 1 = 0\\
\Leftrightarrow 2\cos 6x\cos 4x + 2{\cos ^2}4x = 0\\
\Leftrightarrow 2\cos 4x\left( {\cos 6x + \cos 4x} \right) = 0\\
\Leftrightarrow 2\cos 4x.2\cos 5x\cos x = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 4x = 0\\
\cos 5x = 0\\
\cos x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
4x = \dfrac{\pi }{2} + k\pi \\
5x = \dfrac{\pi }{2} + k\pi \\
x = \dfrac{\pi }{2} + k\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{4}\\
x = \dfrac{\pi }{{10}} + \dfrac{{k\pi }}{5}\\
x = \dfrac{\pi }{2} + k\pi
\end{array} \right.\,\,\left( {k \in Z} \right)
\end{array}\)
