8.146. $\operatorname{tg} x+\operatorname{tg} 50^{\circ}+\operatorname{tg} 70^{\circ}=\operatorname{tg} x \operatorname{tg} 50^{\circ} \operatorname{tg} 70^{\circ}$.
## Solution.

Domain of definition: $\cos x \neq 0$.

From the condition

$\frac{\sin x}{\cos x}+\frac{\sin 50^{\circ}}{\cos 50^{\circ}}+\frac{\sin 70^{\circ}}{\cos 70^{\circ}}-\frac{\sin x \sin 50^{\circ} \sin 70^{\circ}}{\cos x \cos 50^{\circ} \cos 70^{\circ}}=0 \Rightarrow$

$\Rightarrow\left(\sin x \cos 50^{\circ} \cos 70^{\circ}-\sin x \sin 50^{\circ} \sin 70^{\circ}\right)+$

$+\left(\sin 50^{\circ} \cos x \cos 70^{\circ}+\sin 70^{\circ} \cos x \cos 50^{\circ}\right)=0$,

$\sin x\left(\cos 50^{\circ} \cos 70^{\circ}-\sin 50^{\circ} \sin 70^{\circ}\right)+$

$+\cos x\left(\sin 50^{\circ} \cos 70^{\circ}+\cos 50^{\circ} \sin 70^{\circ}\right)=0 \Leftrightarrow$

$\Leftrightarrow \sin x \cos \left(50^{\circ}+70^{\circ}\right)+\cos x \sin \left(50^{\circ}+70^{\circ}\right)=0, \quad \sin x \cos 120^{\circ}+$

$+\cos x \sin 120^{\circ}=0, \sin \left(x+120^{\circ}\right)=0, \quad-\sin \left(x-60^{\circ}\right)=0$.

From this, $x-60^{\circ}=180^{\circ} n, x=60^{\circ}+180^{\circ} n, n \in Z$.

Answer: $x=60^{\circ}+180^{\circ} n, n \in Z$.