Finding the all possible values of $x$ such that $\tan^{-1}(x+1) + \tan^{-1}(x) + \tan^{-1}(x-1) = \tan^{-1}(3)$ Find possible value of $x$ such that
$$\tan^{-1}(x+1) + \tan^{-1}(x) + \tan^{-1}(x-1) = \tan^{-1}(3)$$
Progress: what I did was to consider a case when $x^2 -1 < 1$ $(xy < 1)$ and $3x>-1$ $(xy > -1)$ and then apply $\tan^{-1}(x) \pm tan^{-1}(y)$ identity and got the range.
But is it correct to say we got all the possible value of $x$ or we need to consider all possibilities of $x^2 - 1 > 1$ , $3x < -1$ etc?
As what my book did was just did the first case and then left it without telling what about the solutions from other cases of $tan^{-1}(x) \pm \tan^{-1}(y)$.
 A: I advise you to apply the identity
\begin{align*}
\tan(a + b + c) = \frac{\tan(a) + \tan(b) + \tan(c) - \tan(a)\tan(b)\tan(c)}{1 - \tan(a)\tan(b) - \tan(a)\tan(c) - \tan(b)\tan(c)}
\end{align*}
so that you arrive at the following equation:
\begin{align*}
\frac{3x  - x(x^{2} - 1)}{1 - (x^{2} - 1) - 2x^{2}} = 3 & \Longleftrightarrow \frac{x^{3} - 4x}{3x^{2} - 2} = 3\\\\
& \Longleftrightarrow x^{3} - 9x^{2} - 4x + 6  = 0\\\\
& \Longleftrightarrow (x^{3} + x^{2}) - 10(x^{2} + x) + 6(x + 1) = 0\\\\
& \Longleftrightarrow x^{2}(x^{2} + 1) - 10x(x + 1) + 6(x + 1) = 0\\\\
& \Longleftrightarrow (x^{2} - 10x + 6)(x + 1) = 0
\end{align*}
Hence the solution set is a subset of $S = \{5\pm\sqrt{19},-1\}$.
EDIT
It remains to determine if there are extraneous roots.
In order to do so, you can check out WolframAlpha (for example).
A: You do not need to consider cases.
Think first that you are looking for the zeros of function
$$f(x)=\tan^{-1}(x+1) + \tan^{-1}(x) + \tan^{-1}(x-1) -k$$ the first derivative
$$f'(x)=\frac{1}{1+(x+1)^2}+\frac{1}{1+x^2}+\frac{1}{1+(1-x)^2}=\frac {3 x^4+6 x^2+8 } {\left(x^2+1\right) \left(x^2-2 x+2\right) \left(x^2+2 x+2\right) }$$ is always positive; so, whatever is $k$, there is only one root.
$$k=\tan^{-1}(3)\implies f(0)=-\tan^{-1}(3)<0\quad \text{and}\quad f(1)=\frac{\pi }{4}-\cot ^{-1}(7)>0$$ So, the solution is $\in (0,1)$.
If you do not want to make anything complex, expand $f(x)$ as a series around $x=0$
$$f(x)=-\tan ^{-1}(3)+2 x-\frac{x^3}{6}+O\left(x^5\right)$$ Using the expansion to $O\left(x^3\right)$, the simplest estimate is
$$x_0 \sim \frac 12 \tan ^{-1}(3) =0.624523$$ while the exact solution is $5-\sqrt{19}=0.641101$.
If you already now about series reversion, using the expansion to $O\left(x^5\right)$ gives
$$x=\frac{1}{2} \left(f(x)+\tan ^{-1}(3)\right)+\frac{1}{96} \left(f(x)+\tan
   ^{-1}(3)\right)^3+O\left(\left(f(x)+\tan ^{-1}(3)\right)^5\right)$$ and since we want $f(x)=0$, then the estimate
$$x=\frac{1}{96} \tan ^{-1}(3) \left(48+\tan ^{-1}(3)^2\right)=0.644821$$
A: Since $$\tan\big(\arctan x+\arctan y\big)\equiv\frac{x+y}{1-xy}$$ and $\arctan$ has principal range $\left(-\frac\pi2,\frac\pi2\right),$ thus, for each $(x,y)$ and some $k\in\{-1,0,1\},$ $$\arctan x+\arctan y=\arctan\left(\frac{x+y}{1-xy}\right)+k\pi.$$
Therefore, \begin{align}&\arctan(x+1) + \arctan(x) + \arctan(x-1) = \arctan(3)\\
\implies&\arctan\left(\frac{x^3-4x}{3x^2-2}\right)=\arctan3+n\pi\quad\text{for some }n\in\{-2,-1,\ldots,2\}\\
\implies&\frac{x^3-4x}{3x^2-2}=3\\
\implies&\cdots\\
\implies&x=-1\:\:\text{or}\:\:5\pm\sqrt{19}\end{align}
$5-\sqrt{19}$ is the only non-extraneous solution. Desmos check.
