Đáp án:
\[\left[ \begin{array}{l}
x = \dfrac{{k2\pi }}{3}\\
x = - \dfrac{{2\pi }}{9} + \dfrac{{k2\pi }}{3}
\end{array} \right.\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\cos x.\cos y = \dfrac{1}{2}\left[ {\cos \left( {x + y} \right) + \cos \left( {x - y} \right)} \right]\\
\sin x.\sin y = - \dfrac{1}{2}\left( {\cos \left( {x + y} \right) - \cos \left( {x - y} \right)} \right)\\
\cos \left( {x + y} \right) = \cos x.\cos y - \sin x.\sin y\\
\cos 8x.\cos 5x - \sqrt 3 \sin 3x = 1 - \sin 8x.\sin 5x\\
\Leftrightarrow \dfrac{1}{2}\left[ {\cos \left( {8x + 5x} \right) + \cos \left( {8x - 5x} \right)} \right] - \sqrt 3 \sin 3x = 1 + \dfrac{1}{2}\left[ {\cos \left( {8x + 5x} \right) - \cos \left( {8x - 5x} \right)} \right]\\
\Leftrightarrow \dfrac{1}{2}\left( {\cos 13x + \cos 3x} \right) - \sqrt 3 \sin 3x = 1 + \dfrac{1}{2}\left( {\cos 13x - \cos 3x} \right)\\
\Leftrightarrow \cos 3x - \sqrt 3 \sin 3x = 1\\
\Leftrightarrow \dfrac{1}{2}\cos 3x - \dfrac{{\sqrt 3 }}{2}\sin 3x = \dfrac{1}{2}\\
\Leftrightarrow \cos 3x.\cos \dfrac{\pi }{3} - \sin \dfrac{\pi }{3}.\sin 3x = \cos \dfrac{\pi }{3}\\
\Leftrightarrow \cos \left( {3x + \dfrac{\pi }{3}} \right) = \cos \dfrac{\pi }{3}\\
\Leftrightarrow \left[ \begin{array}{l}
3x + \dfrac{\pi }{3} = \dfrac{\pi }{3} + k2\pi \\
3x + \dfrac{\pi }{3} = - \dfrac{\pi }{3} + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{k2\pi }}{3}\\
x = - \dfrac{{2\pi }}{9} + \dfrac{{k2\pi }}{3}
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)
