`~rai~`
\(1.cos\left(2x+\dfrac{\pi}{6}\right)=\dfrac{\sqrt{3}}{2}\\\Leftrightarrow \left[\begin{array}{I}2x+\dfrac{\pi}{6}=\dfrac{\pi}{6}+k2\pi\\2x+\dfrac{\pi}{6}=-\dfrac{\pi}{6}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}2x=k2\pi\\2x=-\dfrac{\pi}{3}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=k\pi\\x=-\dfrac{\pi}{6}+k\pi.\end{array}\right.\quad(k\in\mathbb{Z})\\2)cos\left(4x-\dfrac{\pi}{3}\right)=\dfrac{1}{3}\\\Leftrightarrow \left[\begin{array}{I}4x-\dfrac{\pi}{3}=arccos\left(\dfrac{1}{3}\right)+k2\pi\\4x-\dfrac{\pi}{3}=-arccos\left(\dfrac{1}{3}\right)+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}4x=arccos\left(\dfrac{1}{3}\right)+\dfrac{\pi}{3}+k2\pi\\4x=-arccos\left(\dfrac{1}{3}\right)+\dfrac{\pi}{3}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{1}{4}arccos\left(\dfrac{1}{3}\right)+\dfrac{\pi}{12}+k\dfrac{\pi}{2}\\x=-\dfrac{1}{4}arccos\left(\dfrac{1}{3}\right)+\dfrac{\pi}{12}+k\dfrac{\pi}{2}.\end{array}\right.\quad(k\in\mathbb{Z})\\3.cos\left(\dfrac{\pi}{5}-x\right)=-2\\\text{Ta có:}-1\le cos\left(\dfrac{\pi}{5}-x\right)\le 1\quad\forall x\\mà\quad -2<-1\\\Rightarrow \text{Phương trình vô nghiệm.}\)
