2
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{
    Sin[Pi/2 + I*Log[2 + Sqrt[3]]],
    Sin[Pi/2 - I*Log[2 + Sqrt[3]]],
    Sin[Pi/2 + I*Log[2 - Sqrt[3]]],
    Sin[Pi/2 - I*Log[2 - Sqrt[3]]]
} // FullSimplify

These will verify to the correct answers,

{2, 2, 2, 2}

So I am hoping to solve this using

Solve[Sin[z] == 2, z, Complexes] // ComplexExpand
Solve[{Sin[z] == 2, Abs[z] < 10}, z] // ComplexExpand
Solve[{Sin[z] == 2, 0 < Re[z] < 10}, z] // ComplexExpand

none gives the answer that has $i\log(2-\sqrt{3})$, which I don't think is equivalent to $i\log(2+\sqrt{3})$ unless I am missing something ?

enter image description here

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3
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Note that Log[2 - Sqrt[3]] == -Log[2 + Sqrt[3]].

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