`~rai~`
\(f)\cos\left(2x+\dfrac{\pi}{6}\right)=\cos\left(x-\dfrac{\pi}{3}\right)\\\Leftrightarrow \left[\begin{array}{I}2x+\dfrac{\pi}{6}=x-\dfrac{\pi}{3}+k2\pi\\2x+\dfrac{\pi}{6}=\dfrac{\pi}{3}-x+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=-\dfrac{\pi}{2}+k2\pi\\3x=\dfrac{\pi}{6}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=-\dfrac{\pi}{2}+k2\pi\\x=\dfrac{\pi}{18}+k\dfrac{2\pi}{3}.\end{array}\right.\quad(k\in\mathbb{Z})\\\text{Vậy S=}\left\{-\dfrac{\pi}{2}+k2\pi;\dfrac{\pi}{18}+k\dfrac{2\pi}{3}\Big|k\in\mathbb{Z}\right\}.\\g)\cos\left(2x+\dfrac{2\pi}{5}\right)+\cos\left(x-\dfrac{\pi}{3}\right)=0\\\Leftrightarrow \cos\left(2x+\dfrac{2\pi}{5}\right)=-\cos\left(x-\dfrac{\pi}{3}\right)\\\Leftrightarrow \cos\left(2x+\dfrac{\pi}{5}\right)=\cos\left(x-\dfrac{\pi}{3}+\pi\right)\\\Leftrightarrow \cos\left(2x+\dfrac{2\pi}{5}\right)=\cos\left(x+\dfrac{2\pi}{3}\right)\\\Leftrightarrow \left[\begin{array}{I}2x+\dfrac{2\pi}{5}=x+\dfrac{2\pi}{3}+k2\pi\\2x+\dfrac{2\pi}{5}=-x-\dfrac{2\pi}{3}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{4\pi}{15}+k2\pi\\3x=-\dfrac{16\pi}{15}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{4\pi}{15}+k2\pi\\x=-\dfrac{16\pi}{45}+k\dfrac{2\pi}{3}.\end{array}\right.\quad(k\in\mathbb{Z})\\\text{Vậy S=}\left\{\dfrac{4\pi}{15}+k2\pi;-\dfrac{16\pi}{45}+k\dfrac{2\pi}{3}\Big|k\in\mathbb{Z}\right\}.\)
