# value of $x$ in Trigonometric equation

Find real $$x(0 in

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

what i try

$$\displaystyle \tan(x+100^\circ)-\tan(x)=\tan(x+50^\circ)+\tan(x-50^\circ)$$

$$\displaystyle \frac{\sin(100^\circ)}{\cos(x+100^\circ)\cos x}=\frac{\sin(2x)}{\cos(x+50^\circ)\cos(x-50^\circ)}$$

$$\displaystyle \frac{\sin(100^\circ)}{\cos(2x+100^\circ)+\cos(100^\circ)}=\frac{\sin(2x)}{\cos(2x)+\cos(100^\circ)}$$

How do i solve further

Let $$\cos100^\circ=a$$ and $$\sin100^\circ=b$$; let $$\cos2x=X$$ and $$\sin2x=Y$$. The final equation can be written $$\frac{b}{aX-bY+a}=\frac{Y}{X+a}$$ that is, $$bX+ab=aXY-bY^2+aY$$ together with $$X^2+Y^2=1$$. This is the intersection between a hyperbola and a circle, so generally a degree 4 equation.
Using the addition theorem for $$\tan$$ you get $$\frac{\tan(x) +\tan(100^\circ)}{1 - \tan(x) \tan(100^\circ)}=\frac{\tan(x) - \tan(50^\circ)}{1 + \tan(x) \tan(50^\circ)}+\tan(x)+\frac{\tan(x) + \tan(50^\circ)}{1 -\tan(x) \tan(50^\circ)}$$ Let $$y= \tan(x)$$, then this is $$\frac{y +\tan(100^\circ)}{1 - y\tan(100^\circ)}=\frac{y- \tan(50^\circ)}{1 + y \tan(50^\circ)}+y+\frac{y + \tan(50^\circ)}{1 -y \tan(50^\circ)}$$ Clearing denominators gives $$0 = - (y +\tan(100^\circ))(1 -y^2 \tan^2(50^\circ))+(y -\tan(50^\circ))(1 -y \tan(50^\circ))(1 - y\tan(100^\circ))\\ +y (1 -y^2 \tan^2(50^\circ))(1 - y\tan(100^\circ)) + (y +\tan(50^\circ))(1 +y \tan(50^\circ))(1 - y\tan(100^\circ))$$
Multiplying out gives $$0 = -\tan^2(50) - 2 y \tan(100) (1 + \tan^2(50)) + y^2 (-1 - 3 \tan^2(50)) + y^4$$
With some help of Wolframalpha, the only positive solution to this fourth order equation is $$\tan x= y \simeq 1.90326$$ or $$x \simeq 62.28^\circ$$ or $$x \simeq 0.346 \pi \simeq 1.087$$ (in radians). I wonder if this "fits" to something special.
The only negative solution is $$\tan x= y \simeq -1.73205$$ which gives $$x \simeq 120^\circ$$. Now this can be verified to be exactly $$x =120^\circ$$ since indeed, using the original equation, $$\tan(220^\circ)=\tan(70^\circ)+\tan(120^\circ)+\tan(170^\circ)$$ holds.