Answer
$\frac{\sqrt {17}-3}{4}$
Work Step by Step
Step 1. Let $u=tan^{-1}x$ and $v=tan^{-1}2x$, we have $tan(u)=x, tan(v)=2x$.
The equation becomes $u+v=\frac{\pi}{4}$
Step 2. Take tangent on both sides of the above, we get $tan(u+v)=tan\frac{\pi}{4}$, use the Addition Formula to obtain $\frac{tan(u)+tan(v)}{1-tan(u)tan(v)}=1$
Step 3. Plug the results from Step 1 in Step 2, we have $\frac{x+2x}{1-2x^2}=1$ which leads to $2x^2+3x-1=0$
Step 4. Solve the above quadratic equation to get $x=\frac{-3\pm\sqrt {3^2+8}}{4}=\frac{-3\pm\sqrt {17}}{4}$ which gives $x=-1.781, 0.281$. As $-1.781\lt -\frac{\pi}{2}$ and it is outside the domain, we get the final answer as $x=0.281$ or $\frac{\sqrt {17}-3}{4}$
