The answer is “”It depends”
If it’s a function, the answer is ##pi/4##
If it’s the fiber, the answer is ##{pi/4 + 2kpi} cup{{3pi}/4 + 2kpi}, k in ZZ##
This is because ##sin## is only locally invertible, so I consider the interval that I prefer which has ##+-pi/2## on the border ##sin^-1:[-pi/2,pi/2] -> [-1,1]##
(if you’re not familiar with this notation don’t worry, it just means that the function doesn’t make any sense for ##xpi/2##)
(If you know something of analysis, this is because in ##+-pi/2## the derivative vanishes and the function is not locally injective, if you don’t, just ignore this and it is because “you see it from the graph”)
So, we knot that the only ##x## we consider such that ##sin(x)=sqrt(2)/2## is ##x=pi/4##
On the other hand, if you want the fiber of ##sqrt(2)/2##, which is the set of all ##x in RR## such that ##sin(x)=sqrt(2)/2##, you got to consider that ##sin(x)## has period ##2pi##, so we know that the only ##x## such that ##x>0, x<2pi## and ##sin(x)=sqrt(2)/2## are ##x=pi/4## and ##x={3pi}/4##, so the fiber is ##{pi/4 + 2kpi} cup{{3pi}/4 + 2kpi}, k in ZZ##
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