`~rai~`
\(2.cos\left(4x-\dfrac{\pi}{3}\right)=\dfrac{1}{3}\\\Leftrightarrow \left[\begin{array}{I}4x-\dfrac{\pi}{3}=arccos\left(\dfrac{1}{3}\right)+k2\pi\\4x-\dfrac{\pi}{3}=-arccos\left(\dfrac{1}{3}\right)+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}4x=arccos\left(\dfrac{\pi}{3}\right)+\dfrac{\pi}{3}+k2\pi\\4x=-arccos\left(\dfrac{1}{3}\right)+\dfrac{\pi}{3}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{1}{4}arccos\left(\dfrac{1}{3}\right)+\dfrac{\pi}{12}+k\dfrac{\pi}{2}\\x=-\dfrac{1}{4}arccos\left(\dfrac{1}{3}\right)+\dfrac{\pi}{12}+k\dfrac{\pi}{2}.\end{array}\right.\quad(k\in\mathbb{Z})\\3)cos\left(\dfrac{\pi}{5}-x\right)=-2\\\text{Ta có:}-1\le\cos{\left(\dfrac{\pi}{5}\right)}\le 1\quad\forall x\\mà\quad -2<-1\\\Rightarrow \text{Phương trình vô nghiệm.}\\4.sin\left(3x+\dfrac{\pi}{3}\right)=-\dfrac{2}{3}\\\Leftrightarrow \left[\begin{array}{I}3x+\dfrac{\pi}{3}=arcsin\left(-\dfrac{2}{3}\right)+k2\pi\\3x+\dfrac{\pi}{3}=\pi-arcsin\left(-\dfrac{2}{3}\right)+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}3x=arcsin\left(-\dfrac{2}{3}\right)-\dfrac{\pi}{3}+k2\pi\\3x=-arcsin\left(-\dfrac{2}{3}\right)+\dfrac{2\pi}{3}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{1}{3}arcsin\left(-\dfrac{2}{3}\right)-\dfrac{\pi}{9}+k\dfrac{2\pi}{3}\\x=-\dfrac{1}{3}arcsin\left(-\dfrac{2}{3}\right)+\dfrac{2\pi}{9}+k\dfrac{2\pi}{3}.\end{array}\right.\quad(k\in\mathbb{Z})\)
