Đk $x \ne \frac{\pi}{6}+\frac{k\pi}{2}$
Do $\tan\left(x+\frac{\pi}{3}\right).\tan\left[\frac{\pi}{2}-\left(x+\frac{\pi}{3}\right)\right]=1$
$\Rightarrow \tan\left(x-\frac \pi6\right).\tan\left(x+\frac \pi3\right)=-1$
$pt\Leftrightarrow \sin^3x.\sin 3x+\cos^3x.\cos 3x=\frac 18$
$\Leftrightarrow\sin^2x\left(\cos 2x-\cos 4x\right)+\cos^2x\left(\cos 2x+\cos 4x\right)=\frac 14$
$\Leftrightarrow (1-\cos 2x)(\cos 2x-\cos 4x)+(1+\cos 2x)(\cos 2x+\cos 4x)=\frac 12$
$\Leftrightarrow (1-t)(t-2t^2+1)+(1+t)(t+2t^2-1)=\frac 12$
$\Leftrightarrow 4t^3=\frac 12\Leftrightarrow t=\frac 12\Leftrightarrow \cos 2x=\frac 12\Leftrightarrow \left[ \begin{array}{l} x=\frac \pi6+k\pi\\ x=-\frac \pi6+k\pi \end{array} \right.$
Kết hợp đk, pt có họ nghiệm $x=-\frac \pi6+k\pi,k \in \mathbb Z$