# Solving equation involving inverse trig functions

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

Solve for $$x$$ if $$x∈\Bbb R^{+}$$

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

• Welcome to MSE! What have you tried and what do you know about $\tan$? In particular, do you know an expression for $\tan(a+b)$? Commented Jun 27, 2018 at 13:30
• Apply $\tan$ to both sides and exploit $\tan(a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}$. What do you get? Commented Jun 27, 2018 at 13:30
• You may also consider that $x=\frac{1}{2}$ is a trivial solution and the LHS is an increasing function, so the trivial solution is the only solution. Commented Jun 27, 2018 at 13:32

Hint:

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

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

Hint:

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

• sorry. maybe i did some mistake. Commented Jun 27, 2018 at 14:01
• no. sorry. it was a mistake. i am ready to upvote this answer. Commented Jun 27, 2018 at 14:03
• but i am not allowed to upvote it any longer. sorry for downvoting. i am using stack exchange after a long time. Commented Jun 27, 2018 at 14:07