# $\arcsin(x) = \arctan(2x)$

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}

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?

so, if Arcsin(x) = arctan(2x)

then x is the sin of whatever Arcsin(x) is, so therefore the tan of the thing would be

x / sqrt(1 - x^2)

so, we can then drop the arctan

x^ 2 / (1 - x^2) = 4x^2

x^2 = 1 - 1/4 = 3/4

x = sqrt(3/4)

• Are there $\pm$ to consider? – Acccumulation Sep 24 '18 at 16:38
• what do you think? Go and investigate it. – Cato Sep 25 '18 at 8:57

The LHS is defined only for $$x\in[-1,1]$$ and both functions are odd, hence it is enough to look for solutions in $$[0,1]$$. $$x=0$$ is obviously one of them, and there is an extra solution in $$(0,1]$$, since over such interval both functions are increasing, $$\arcsin(x)$$ is convex, $$\arctan(2x)$$ is concave and $$\arcsin(1)>\arctan(2)$$.
In order to solve $$\tan\arcsin(x) = 2x$$ it is enough to solve $$\frac{x}{\sqrt{1-x^2}} = 2x$$ or $$\sqrt{1-x^2} = \frac{1}{2}$$ hence the wanted solutions are $$x\in\left\{-\frac{\sqrt{3}}{2},0,\frac{\sqrt{3}}{2}\right\}$$.