Supposedly easy arcsine problem

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Homework Statement



find x if sin(x)=1/5

Homework Equations



x=arcsin(1/5)= ??something in closed form maybe??

The Attempt at a Solution



This actually isn't my homework. A student I am tutoring in differential calculus came to me with this problem and I am convinced it is a typo. Just for certainty, is there a closed form for this? I started messing around with complex numbers and series... so i thought it might be time to ask here.
Thanks!
 
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so i just want to confirm that it is not really a fair and square Cal 1 problem...
 
I doubt you are going to find a neater form for that. It's arcsin(1/5).
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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