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

Solve for $x$ if $x∈\Bbb R^{+}$

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

Sebastiano
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StacyMeghan
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  • Welcome to MSE! What have you tried and what do you know about $\tan$? In particular, do you know an expression for $\tan(a+b)$? – Bill O'Haran Jun 27 '18 at 13:30
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    Apply $\tan$ to both sides and exploit $\tan(a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}$. What do you get? – Jack D'Aurizio Jun 27 '18 at 13:30
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    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. – Jack D'Aurizio Jun 27 '18 at 13:32

2 Answers2

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

lab bhattacharjee
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Hint:

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

Bernard
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