Đáp án:
$x = \dfrac{\pi}{3} + k\pi \quad (k \in \Bbb Z)$
Giải thích các bước giải:
$\begin{array}{l}2\tan x + \cot x = \sqrt3 + \dfrac{2}{\sin2x} \qquad (*)\\ ĐK: \quad \sin2x \ne 0 \Leftrightarrow x \ne k\dfrac{\pi}{2} \quad (k \in \Bbb Z)\\ (*) \Leftrightarrow 2.\dfrac{\sin x}{\cos x} + \dfrac{\cos x}{\sin x} = \sqrt3 + \dfrac{2}{2\sin x\cos x}\\ \Leftrightarrow \dfrac{2\sin^2x + \cos^2x}{\sin x\cos x} = \sqrt3 + \dfrac{1}{\sin x\cos x}\\ \Leftrightarrow 2\sin^2x + \cos^2x = \sqrt3\sin x\cos x + 1\\ \Leftrightarrow 2.\dfrac{1 - \cos2x}{2} + \dfrac{1 + \cos2x}{2} = \sqrt3\sin x\cos x + 1\\ \Leftrightarrow 2 - 2\cos2x + 1 + \cos2x = 2\sqrt3\sin x\cos x + 2\\ \Leftrightarrow \sqrt3\sin2x + \cos2x = 1\\ \Leftrightarrow \dfrac{\sqrt3}{2}\sin2x + \dfrac{1}{2}\cos2x = \dfrac{1}{2}\\ \Leftrightarrow \cos\dfrac{\pi}{6}\sin2x + \sin\dfrac{\pi}{6}\cos2x = \sin\dfrac{\pi}{6}\\ \Leftrightarrow \sin\left(2x + \dfrac{\pi}{6}\right) = \sin\dfrac{\pi}{6}\\ \Leftrightarrow \left[\begin{array}{l}2x + \dfrac{\pi}{6} = \dfrac{\pi}{6} + k2\pi\\2x + \dfrac{\pi}{6} = \dfrac{5\pi}{6} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}2x= k2\pi\\2x = \dfrac{2\pi}{3} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x= k\pi\quad (loại)\\x = \dfrac{\pi}{3} + k\pi\end{array}\right.\qquad (k \in \Bbb Z)\\ Vậy\,\,x = \dfrac{\pi}{3} + k\pi \quad (k \in \Bbb Z)\end{array}$
