Calculus me solve this simple problem (I forgot how to): arcsin(4/5) = ?

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Homework Statement



how do I solve this

arcsin(4/5)

note that I am not looking for about 53 degrees

I believe I'm suppose to solve euler's formula for x

i.e.
sin (x) = (e^(ix) - e^(-ix))/(2i)
where x is in radians
hence I would do something like this

sin (x) = (e^(ix) - e^(-ix))/(2i) = 4/5

(e^(ix) - e^(-ix))/(2i) = 4/5
solve the equation above for x

this is were I need help if somebody could just show me quickly how to do this that would be great!

If I remeber correctly I need to use cis(x) or something right?


Homework Equations





The Attempt at a Solution

 
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ok now this is an algebra 2 problem... I have goten far down to hear

e^(ix) = (4i +/- 3)/5

now how do I solve for x I feel really stupid now as this is algebra 2...

can't use natural log right or the common log as

ln( e^(ix) )
does not equal ix

so what do I do?
 


Wait apparently I'm suppose to do something involving

log(z) = ln|z| + i arg(z) for complex numbers z

or something correct?
 


Try it and see.
 
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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