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?