$\sqrt3\sin x +\cos x =\dfrac{1}{\cos x}\quad (*)$

ĐKXĐ: $\cos x \ne 0 \Leftrightarrow x \ne \dfrac{\pi}{2} + k\pi$

$(*)\Leftrightarrow \sqrt3\sin x\cos x + \cos^2x - 1 = 0$

$\Leftrightarrow \sqrt3\sin x\cos x - \sin^2x = 0$

$\Leftrightarrow \sin x\left(\sqrt3\cos x -\sin x\right)= 0$

$\Leftrightarrow \sin x\left(\dfrac{\sqrt3}{2}\cos x -\dfrac12\sin x\right)= 0$

$\Leftrightarrow \sin x\cos\left(x + \dfrac{\pi}{6}\right)= 0$

$\Leftrightarrow \left[\begin{array}{l}\sin x = 0\\\cos\left(x + \dfrac{\pi}{6}\right)= 0\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}x = k\pi\\x +\dfrac{\pi}{6}=\dfrac{\pi}{2} + k\pi\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}x = k\pi\\x = \dfrac{\pi}{3} + k\pi\end{array}\right.\quad (k\in\Bbb Z)$

Vậy phương trình có tập nghiệm

$S =\left\{k\pi,\ \dfrac{\pi}{3} + k\pi\ \Bigg|\ k\in\Bbb Z\right\}$