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The answer is:

##arcsin(sin(7pi/6))=-pi/6##.

The range of a function ##arcsin(x)## is by definition

##-pi/2<=arcsin(x)<=pi/2##
It means that we have to find an angle ##alpha## that lies between ##-pi/2## and ##pi/2## and whose ##sin(alpha)## equals to a ##sin(7pi/6)##.
From trigonometry we know that
##sin(phi+pi)=-sin(phi)##
for any angle ##phi##.
This is easy to see if use the definition of a sine as an ordinate of the end of a radius in the unit circle that forms an angle ##phi## with the X-axis (counterclockwise from the X-axis to a radius).
We also know that sine is an odd function that is
##sin(-phi)=-sin(phi)##.
We will use both properties as follows:
##sin(7pi/6)=sin(pi/6+pi)=-sin(pi/6)=sin(-pi/6)##
As we see the angle ##alpha=-pi/6## fits our conditions. It is in the range from ##-pi/2## to ##pi/2## and its sine equals to ##sin(7pi/6)##. Therefore it's a correct answer to a problem.

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