$\begin{array}{l}1a)\,\,\cot(x - \pi) = \dfrac{1}{\sqrt3}\\ \Leftrightarrow \cot x = \dfrac{1}{\sqrt3} \qquad (*)\\ ĐKXĐ:\, \sin x \ne 0 \Leftrightarrow x \ne n\pi \quad (k \in \Bbb Z)\\ (*) \Leftrightarrow \cot x = \cot\dfrac{\pi}{3}\\ \Leftrightarrow x = \dfrac{\pi}{3} +k\pi \quad (k \in \Bbb Z)\\ 1b)\,\,\sqrt3\sin2x -2 = \cos2x\\ \Leftrightarrow \dfrac{\sqrt3}{2}\sin2x - \dfrac{1}{2}\cos2x = 1\\ \Leftrightarrow \sin2x\cos\dfrac{\pi}{6} - \cos2x\sin\dfrac{\pi}{6} = 1\\ \Leftrightarrow \sin\left(2x - \dfrac{\pi}{6}\right) = 1\\ \Leftrightarrow 2x - \dfrac{\pi}{6} = \dfrac{\pi}{2} +k2\pi\\ \Leftrightarrow 2x = \dfrac{2\pi}{3} + k2\pi\\ \Leftrightarrow x = \dfrac{\pi}{3} +k\pi\quad (k \in \Bbb Z)\\ 2) \,\,\sin x + \sin2x = \cos x +2\cos^2x\\ \Leftrightarrow \sin x + 2\sin x\cos x = \cos x +2\cos^2x\\ \Leftrightarrow \sin x(1 + 2\cos x) = \cos x(1 + 2\cos x)\\ \Leftrightarrow (1+2\cos x)(\sin x - \cos x) = 0\\ \Leftrightarrow \left[\begin{array}{l}2\cos x + 1 = 0\\\sin x - \cos x = 0\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}\cos x = - \dfrac{1}{2}\\\sin\left(x - \dfrac{\pi}{4}\right) = 0\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = - \dfrac{2\pi}{3} + k2\pi\\x = \dfrac{2\pi}{3} + k2\pi\\x = \dfrac{\pi}{4} + k\pi\end{array}\right.\quad (k \in \Bbb Z)\\ +) \quad k = 0\\ \Rightarrow \left[\begin{array}{l}x = - \dfrac{2\pi}{3}\\x = \dfrac{2\pi}{3}\\x = \dfrac{\pi}{4} \end{array}\right.\\ +) \quad k = -1\\ \Leftrightarrow \left[\begin{array}{l}x = - \dfrac{8\pi}{3}\\x = -\dfrac{4\pi}{3}\\x = -\dfrac{3\pi}{4}\end{array}\right.\\ \text{So sánh các nghiệm âm, ta được:}\\ -\dfrac{2\pi}{3}; \, - \dfrac{3\pi}{4};\, - \dfrac{4\pi}{3}\\ \text{lần lượt là 3 nghiệm âm lớn nhất} \end{array}$
