`~rai~`
\(a)cos\left(x-\dfrac{\pi}{6}\right)=\sqrt{\dfrac{5}{3}}\\Vì\quad -1\le cos\left(x-\dfrac{\pi}{6}\right)\le 1\quad mà\quad\sqrt{\dfrac{5}{3}}>1\\\Rightarrow \text{Phương trình vô nghiệm.}\\b)cos\left(2x+\dfrac{\pi}{6}\right)=0\\\Leftrightarrow 2x+\dfrac{\pi}{6}=\dfrac{\pi}{2}+k\pi\\\Leftrightarrow 2x=\dfrac{\pi}{3}+k\pi\\\Leftrightarrow x=\dfrac{\pi}{6}+k\dfrac{\pi}{2}.(k\in\mathbb{Z})\\c)cos\left(\dfrac{\pi}{5}-x\right)=-1\\\Leftrightarrow \dfrac{\pi}{5}-x=\pi+k2\pi\\\Leftrightarrow x=-\dfrac{4\pi}{5}-k2\pi.(k\in\mathbb{Z})\\d)cos\left(4x-\dfrac{\pi}{3}\right)=-\dfrac{1}{2}\\\Leftrightarrow cos\left(4x-\dfrac{\pi}{3}\right)=cos\left(\dfrac{2\pi}{3}\right)\\\Leftrightarrow \left[\begin{array}{I}4x-\dfrac{\pi}{3}=\dfrac{2\pi}{3}+k2\pi\\4x-\dfrac{\pi}{3}=-\dfrac{2\pi}{3}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}4x=\pi+k2\pi\\4x=-\dfrac{\pi}{3}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{\pi}{4}+k\dfrac{\pi}{2}\\x=-\dfrac{\pi}{12}+k\dfrac{\pi}{2}.\end{array}\right.\quad(k\in\mathbb{Z})\)
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