$Dk: x khác \pi +k2\pi và \frac{\pi }{2}+k\pi (k\in Z)$
$\Leftrightarrow 1+tan^{2}x-(1-2sin^{2}\frac{x}{2}-2\sin \frac{x}{2}\cos \frac{x}{2}.\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}})=\frac{\sin( x-\frac{\pi }{6})+\sin (x+\frac{\pi }{6})}{\cos x}$
$tan^{2}x=\frac{2\sin x.\cos (\frac{-\pi }{6})}{\cos x}\Leftrightarrow tan^{2}x=\sqrt{3}\tan x$
$\Leftrightarrow \tan x=0 hoặc \tan x=\sqrt{3}$
$\Leftrightarrow x=k2\pi hoặc x=\frac{\pi }{3}+k\pi $