Trigonometry equation with arctan Solve the following equation: $\arctan x + \arctan (x^2-1) = \frac{3\pi}{4}$.
What I did
Let  $\arctan x = \alpha, \arctan(x^2-1) = \beta$, $\qquad\alpha+\beta = \frac{3\pi}{4}$
$\tan(\alpha+\beta) = \tan(\frac{3\pi}{4}) = -1$
$$\frac{\tan\alpha + tan\beta}{1-\tan\alpha\tan\beta} = \frac{x+x^2-1}{1-x(x^2-1)} = -1$$
$\begin{align}
x^2+x-1 &= -(1-(x^3-x)) = -1+x^3-x \\
\iff x^2 + x &= x^3-x \\
\iff x(x+1) &= x(x^2-1) \qquad\implies \boxed{x_1 = 0}\\
\implies x+1 &= x^2-1 \\
\iff x^2-x-2 &= 0 \\
\end{align}$
$\therefore x_1 = 0,\quad x_2 = 2,\quad x_3 = -1$
However, the equation only works for $x=2$.
I wonder
Did I do this in an efficient manner? Is there any easy way to find $x$ where there's no fake solutions?

Liek Show that $2\tan^{-1}(2) = \pi - \cos^{-1}(\frac{3}{5})$, 
$$\arctan x+\arctan(x^2-1)=\begin{cases} \arctan\left(\dfrac{x+x^2-1}{1-x(x^2-1)}\right) &\mbox{if } x(x^2-1)<1 \\\pi+ \arctan\left(\dfrac{x+x^2-1}{1-x(x^2-1)}\right) & \mbox{if } x(x^2-1)> 1. \end{cases} $$ 
Now, $-\dfrac\pi2\le\arctan(z)\le\dfrac\pi2$
$\implies-\dfrac\pi2\le\arctan x+\arctan(x^2-1)\le\dfrac\pi2$ if $x(x^2-1)<1$ which is true if $x=0,-1$
Then $\arctan x+\arctan(x^2-1)\ne\dfrac{3\pi}4$
