$\underline{\textbf{Solution 1}}$ In the following identity (see # 3851) $$\sin\alpha + \sin\beta + \sin\gamma - \sin(\alpha + \beta+\gamma) = 4\sin\frac{\alpha+\beta}{2}\sin\frac{\beta+\gamma}{2}\sin\frac{\gamma+\alpha}{2}$$ Setting $\alpha=x$, $\beta=x+120^\circ$, and $\gamma=x-120^\circ$ leads \begin{align} \sin x +\sin(x+120^\circ)+\sin(x-120^\circ) + \sin 3x &=4\sin(x+60^\circ)\sin x \sin (x-60^\circ) \end{align} It can be shown that $\sin x + \sin (x+120^\circ) + \sin(x-120^\circ) = 0$ (see # 3860) . Hence the to-be-claimed identity follows.

$\underline{\textbf{Solution 2}}$ The key is to understand where angle $60^\circ$ comes from. We note that $\sin^2 60^\circ = \frac{3}{4}$, which links to the ratio of $3$ and $4$. Therefore, by # 2256, \begin{align} \sin 3\theta &= 3\sin\theta - 4\sin^3\theta\\ &= 4\sin\theta\cdot\Big(\frac{3}{4}-\sin^2\theta\Big)\\ &= 4\sin\theta\cdot\Big(\sin^2 60^\circ - \sin^2\theta)\\ &=4\sin\theta\sin(60^\circ + \theta)\sin(60^\circ - \theta) \end{align} Formula $\cos(3\theta)$ can be solved similarly. Then dividing $\sin (3\theta)$ by $\cos(3\theta)$ will yield the tangent formula immediately.