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Find real $x(0<x<180^\circ)$ 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

Help me please

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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.

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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.

Though this treatment is sort of ugly, it has the advantage that it shows the only two solutions to the problem.

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