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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}

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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?

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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)

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  • $\begingroup$ Are there $\pm$ to consider? $\endgroup$ – Acccumulation Sep 24 '18 at 16:38
  • $\begingroup$ what do you think? Go and investigate it. $\endgroup$ – Cato Sep 25 '18 at 8:57
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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 $x\in\left\{-\frac{\sqrt{3}}{2},0,\frac{\sqrt{3}}{2}\right\}$.

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