Obviously no regular angle, no real number, has a sine of #2#. We're getting in the realm of complex numbers.

Euler's Formula is valid for complex numbers too:

# e^{iz} = cos z + i sin z #

We want to know about #e^{i(-z)} = e^{-iz}.# If #z# is real that would be the conjugate, but it's not for complex #z#. But even for complex #z# we want cosine to be even, #cos z = cos(-z),# and sine to be odd, #sin z = -sin(-z).# So

#e^{-iz} = e^{i(-z)} = cos(-z) + i sin(-z) = cos z - i sin z #

Subtracting,

# e^{iz} - e^{-iz} = 2 i sin z #

#sin z = 1/{2i}(e^{iz} - e^{-iz}) #

From here out we'll use that essentially as the definition of #sin z.#

It's actually easier to get a relatively nice closed form for the inverse sine of a real number greater than one than of a regular sine.

We want to solve

#sin z = a # for real #a>1.#

# 1/{2i}(e^{iz} - e^{-iz}) = a #

Let #y=e^{iz}.# Then #e^{-iz}=1/y.#

#y - 1/y = 2i a #

#y^2 - 2i a y - 1 = 0#

I call it the **Shakespeare Quadratic Formula** (#2b# or #-2b#): #x^2-2bx+c# has zeros #x=b pm sqrt{b^2-c}.#

#y = ia pm sqrt{(-ia)^2+1} = ia pm sqrt{1-a^2} #

Since #a>1# the radicand is negative. Let's make that explicit.

#y = ia pm sqrt{ (a^2-1) (-1) } = i (a pm sqrt{a^2-1} ) #

#e^{iz} = i (a pm sqrt{a^2-1} ) #

Since #i=e^{i pi/2}# and #e^{2pi k i} = 1 quad # integer #k,#

#e^{iz} = e^{i pi /2 } e^{2pi k i} e^{ln (a pm sqrt{a^2-1} ) } #

#iz = i pi/2 + 2pi k i + ln ( a pm sqrt{a^2-1} ) #

#z = (pi/2 + 2 pi k) - i ln ( a pm sqrt{a^2-1} ) #

That's the general solution to #sin z = a # for real #a>1.# There are a few things to note about it. Let's look at the two imaginary parts first.

# (a + sqrt{a^2 - 1} ) (a - sqrt{a^2 - 1 }) = a^2 - (a^2-1) =1 #

# a - sqrt{a^2 - 1 } = 1/{ a + sqrt{a^2 - 1} } #

#ln (a - sqrt{a^2 - 1 } ) = -ln ( a + sqrt{a^2 - 1} ) #

#z = (pi/2 + 2 pi k) pm i ln ( a - sqrt{a^2-1} ) #

The two imaginary parts are negations of each other! That means the #z#s come in complex conjugate pairs.

Considering #k=0#, when we add the two roots together we'll get #pi.# In other words those two roots, even though complex, are supplementary angles who share the same sine, just like real angles. Just like real angles, we can add or subtract #2pi# as many times as we like and get another number with the same sine.

OK, let's plug in #a=2# for the big finish.

#arcsin 2 = (pi/2 + 2pi k ) pm i ln(2 - sqrt{3}) quad# integer #k#