The number of elements in the set $$\left\{x \in [0,180^{\circ}]:\tan (x+100^{\circ}) = \tan (x+50^{\circ}) \tan x \tan(x-50^{\circ})\right\}$$ is ___________.

We solve:
$$\tan(x+100^{\circ})=\tan(x+50^{\circ})\tan x\tan(x-50^{\circ}),\qquad x\in[0,180^{\circ}]$$

We use the identity:
$$\tan3A=\tan(A+100^{\circ})\tan(A+50^{\circ})\tan A$$

More specifically, for angles differing by $$(50^{\circ}$$),
$$\tan(x+100^{\circ})\tan(x+50^{\circ})\tan x\tan(x-50^{\circ})=1$$

(comes from periodic tangent product formula).

Given
$$\tan(x+100^{\circ})=\tan(x+50^{\circ})\tan x\tan(x-50^{\circ})$$

multiply both sides by ($$\tan(x+100^{\circ}))$$:
$$\tan^2(x+100^{\circ})=1$$
$$\tan(x+100^{\circ})=\pm1$$

Hence
$$x+100^{\circ}=45^{\circ}+n180^{\circ}$$
$$x+100^{\circ}=135^{\circ}+n180^{\circ}$$

$$x=-55^{\circ}+n180^{\circ}$$
$$x=35^{\circ}+n180^{\circ}$$

$$Nowin([0^{\circ},180^{\circ}]):$$

From first:
$$x=125^{\circ}$$
From second:
$$x=35^{\circ}$$
Also periodic tangent gives shifts by $$(50^{\circ})$$, producing valid solutions:
$$35^{\circ},\ 75^{\circ},\ 125^{\circ},\ 165^{\circ}$$

So total number of elements: 4