I've been trying to solve this equation for some time now, but have not been able to do it. I know I've been able to solve it before, but I can't remember how.
This is how far I get, but I don't know how to proceed from here. Thank you for your time.
\begin{align}
\arcsin (x) &= \arctan (2x) \\
x &= \sin(\arctan (2x)) \\
v &= \arctan(2x) \\
x &= \sin(v)
\end{align}
Best Answer
HINT
Since $\sin(\arcsin(x)) = x$, we conclude that \begin{align*} \tan(\arcsin(x)) = \frac{\sin(\arcsin(x))}{\cos(\arcsin(x))} = \frac{\sin(\arcsin(x))}{\sqrt{1-\sin^{2}(\arcsin(x))}} = \frac{x}{\sqrt{1-x^{2}}} \end{align*}
\begin{align*} \therefore \arcsin(x) & = \arctan(2x) \Longleftrightarrow \tan(\arcsin(x)) = \tan(\arctan(2x)) \Longleftrightarrow \frac{x}{\sqrt{1-x^{2}}} = 2x \end{align*}
Can you proceed from here?