Find the solutions of $x$ such that $\arcsin x=2\arctan x.$
My solution goes like this:
We have, $\arcsin x=2\arctan x.$ Now, $2\arctan x=\arctan(\frac{2x}{{1-x^2}})=\arcsin(\frac{2x}{1+x^2}).$ Now, $$\arctan(\frac{2x}{1-x^2})=\arcsin (\frac{2x}{1+x^2})=\arcsin x\implies x=(\frac{2x}{1+x^2}).$$ Thus, the possible values of $x$ are $0,1,-1.$
However, if we solve this, using this approach, we find only one solution :
We have, $\arcsin x=2\arctan x.$ Now, $2\arctan x=\arctan(\frac{2x}{1-x^2})$ and $\arcsin x=\arctan (\frac{x}{\sqrt{1-x^2}}).$ Thus, $\arctan (\frac{x}{\sqrt{1-x^2}})=\arctan(\frac{2x}{1+x^2})\implies (\frac{2x}{{1-x^2}})=(\frac{x}{\sqrt{1-x^2}}).$ Now, $x=0$ is the only solution of it. For if, $x\neq 0$, then $1-x^2=4,$ a contradiction.
I don't get what's wrong with this second approach ? Which of the two methods are valid ? If one is invalid, then why is it so ? I am not quite getting it…
Best Answer
The equality $2\arctan x=\arctan\left(\frac{2x}{1-x^2}\right)$ is valid only for $x \in (-1,1)$. It's clear that it's not valid, for example, for $x=1$ since the RHS is undefined.