Question

$$y= \arctan \sqrt{ 4 { x }^{ 2 } -1 }$$

Answer

x=sqrt(y/(4*q*r*t*arctan(s))+1/4),-sqrt(y/(4*q*r*t*arctan(s))+1/4)

Solution


Regroup terms.
\[y=qrt(4{x}^{2}-1)\tan^{-1}{(s)}\]
Divide both sides by \(q\).
\[\frac{y}{q}=rt(4{x}^{2}-1)\tan^{-1}{(s)}\]
Divide both sides by \(r\).
\[\frac{\frac{y}{q}}{r}=t(4{x}^{2}-1)\tan^{-1}{(s)}\]
Simplify  \(\frac{\frac{y}{q}}{r}\)  to  \(\frac{y}{qr}\).
\[\frac{y}{qr}=t(4{x}^{2}-1)\tan^{-1}{(s)}\]
Divide both sides by \(t\).
\[\frac{\frac{y}{qr}}{t}=(4{x}^{2}-1)\tan^{-1}{(s)}\]
Simplify  \(\frac{\frac{y}{qr}}{t}\)  to  \(\frac{y}{qrt}\).
\[\frac{y}{qrt}=(4{x}^{2}-1)\tan^{-1}{(s)}\]
Divide both sides by \(\tan^{-1}{(s)}\).
\[\frac{\frac{y}{qrt}}{\tan^{-1}{(s)}}=4{x}^{2}-1\]
Simplify  \(\frac{\frac{y}{qrt}}{\tan^{-1}{(s)}}\)  to  \(\frac{y}{qrt\tan^{-1}{(s)}}\).
\[\frac{y}{qrt\tan^{-1}{(s)}}=4{x}^{2}-1\]
Add \(1\) to both sides.
\[\frac{y}{qrt\tan^{-1}{(s)}}+1=4{x}^{2}\]
Divide both sides by \(4\).
\[\frac{\frac{y}{qrt\tan^{-1}{(s)}}+1}{4}={x}^{2}\]
Simplify  \(\frac{\frac{y}{qrt\tan^{-1}{(s)}}+1}{4}\)  to  \(\frac{\frac{y}{qrt\tan^{-1}{(s)}}}{4}+\frac{1}{4}\).
\[\frac{\frac{y}{qrt\tan^{-1}{(s)}}}{4}+\frac{1}{4}={x}^{2}\]
Simplify  \(\frac{\frac{y}{qrt\tan^{-1}{(s)}}}{4}\)  to  \(\frac{y}{4qrt\tan^{-1}{(s)}}\).
\[\frac{y}{4qrt\tan^{-1}{(s)}}+\frac{1}{4}={x}^{2}\]
Take the square root of both sides.
\[\pm \sqrt{\frac{y}{4qrt\tan^{-1}{(s)}}+\frac{1}{4}}=x\]
Switch sides.
\[x=\pm \sqrt{\frac{y}{4qrt\tan^{-1}{(s)}}+\frac{1}{4}}\]