0
First, take the inverse sine of both sides of the equation.
That gives you x = sin-1(6), which is sadly undefined...in reality, but who needs that!
It can be proven that sin-1(x) = -i*log[i*x + √(1-x2)]
So in this case:
= -i*log[i*6 + √(1-36)] = -i*log[6*i + √(-35)]
= -i*log(11.916*i)
= 1.57 - 2.48*i
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