Giải đáp: \(S = \left\{- \dfrac{\pi}{30} + \dfrac{k2\pi}{5};\ – \dfrac{\pi}{10} + \dfrac{k2\pi}{5}\ \Bigg|\ k\in\Bbb Z\right\}\) Lời giải và giải thích chi tiết: \(\begin{array}{l}\quad \cos6x\cos x- \sqrt3\sin5x=1-\sin6x\sin x\\\Leftrightarrow (\cos6x\cos x+\sin6x\sin x)- \sqrt3\sin5x=1\\\Leftrightarrow \cos5x – \sqrt3\sin5x=1\\\Leftrightarrow \dfrac12\cos5x – \dfrac{\sqrt3}{2}\sin5x = \dfrac12\\\Leftrightarrow \cos\left(5x + \dfrac{\pi}{3}\right) = \cos\dfrac{\pi}{6}\\\Leftrightarrow \left[\begin{array}{l}5x + \dfrac{\pi}{3} = \dfrac{\pi}{6} + k2\pi\\5x + \dfrac{\pi}{3} = -\dfrac{\pi}{6} + k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{l}x = – \dfrac{\pi}{30} + \dfrac{k2\pi}{5}\\x = – \dfrac{\pi}{10} + \dfrac{k2\pi}{5}\end{array}\right.\quad (k\in\Bbb Z)\\\text{Vậy}\ S = \left\{- \dfrac{\pi}{30} + \dfrac{k2\pi}{5};\ – \dfrac{\pi}{10} + \dfrac{k2\pi}{5}\ \Bigg|\ k\in\Bbb Z\right\}\end{array}\)
\quad \cos6x\cos x- \sqrt3\sin5x=1-\sin6x\sin x\\
\Leftrightarrow (\cos6x\cos x+\sin6x\sin x)- \sqrt3\sin5x=1\\
\Leftrightarrow \cos5x – \sqrt3\sin5x=1\\
\Leftrightarrow \dfrac12\cos5x – \dfrac{\sqrt3}{2}\sin5x = \dfrac12\\
\Leftrightarrow \cos\left(5x + \dfrac{\pi}{3}\right) = \cos\dfrac{\pi}{6}\\
\Leftrightarrow \left[\begin{array}{l}5x + \dfrac{\pi}{3} = \dfrac{\pi}{6} + k2\pi\\5x + \dfrac{\pi}{3} = -\dfrac{\pi}{6} + k2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = – \dfrac{\pi}{30} + \dfrac{k2\pi}{5}\\x = – \dfrac{\pi}{10} + \dfrac{k2\pi}{5}\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy}\ S = \left\{- \dfrac{\pi}{30} + \dfrac{k2\pi}{5};\ – \dfrac{\pi}{10} + \dfrac{k2\pi}{5}\ \Bigg|\ k\in\Bbb Z\right\}
\end{array}\)