`~rai~`
$\begin{array}{1}cos(2x+1)=\dfrac{1}{2}\\\Leftrightarrow \left[\begin{array}{1}x_1=-\dfrac{1}{2}+\dfrac{\pi}{6}+k\pi\\x_2=-\dfrac{1}{2}-\dfrac{\pi}{6}+k\pi(k\in\mathbb{Z})\end{array}\right.\\TH1:-\pi<x_1<\pi\\\Leftrightarrow -\pi<-\dfrac{1}{2}+\dfrac{\pi}{6}+k\pi<\pi\\\Leftrightarrow -1<k<1\\\text{Vì k}\in\mathbb{Z}\Rightarrow k=0\Leftrightarrow x=-\dfrac{1}{2}+\dfrac{\pi}{6}.(1)\\TH2:-\pi<x_2<\pi\\\Leftrightarrow -\pi<-\dfrac{1}{2}-\dfrac{\pi}{6}+k2\pi<\pi\\\Leftrightarrow -0,67<k<1,33\\\text{Vì k}\in\mathbb{Z}\Rightarrow k=0;k=1\\\Leftrightarrow x=-\dfrac{1}{2}-\dfrac{\pi}{6};x=-\dfrac{1}{2}+\dfrac{5\pi}{6}.(2)\\\text{Từ (1) và (2)}\Leftrightarrow \left[\begin{array}{1}x=-\dfrac{1}{2}+\dfrac{\pi}{6}\\x=-\dfrac{1}{2}+\dfrac{\pi}{6}\\x=-\dfrac{1}{2}+\dfrac{5\pi}{6}.\end{array}\right. \end{array}$
