<s>[INST] value of $x$ in Trigonometric equation 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 
 [/INST] 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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