Solving equation involving inverse trig functions

Question

I'd like some help getting this question on its way, any help would be greatly appreciated.

Solve for $x$ if $x∈\Bbb R^{+}$

$$\tan ^{-1}(2x+1)+\tan ^{-1}(2x-1)=\tan ^{-1}2$$

Answer 1

Hint:

Equal the tangents of both sides and use the addition formula for the tangent. You should obtain the quadratic equation $$2x^2+x-1=0.$$

Answer 2

Hint:

Using my answer here Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$,

$$\arctan(2x+1)+\arctan(2x-1)=\begin{cases} \arctan\frac{2x}{1-2x^2} &\mbox{if } (2x+1)(2x-1)<1\\ \pi+\arctan\frac{2x}{1-2x^2} & \mbox{if }(2x+1)(2x-1)>1\\\text{sign}(x)\cdot\dfrac\pi2 & \mbox{if } (2x+1)(2x-1)=1\end{cases}$$

Now $\dfrac\pi2>\arctan(2)>\arctan(\sqrt3)=\dfrac\pi3$