First consider a right angled triangle with sides #1#, #sqrt(3)# and #2#. We can tell that it is right angled because is satisfies:
#1^2+sqrt(3)^2 = 1 + 3 = 4 = 2^2#
Notice that this is one half of an equilateral triangle with side #2#.
The angles of that equilateral triangle are all #pi/3#
So the smallest angle of the right angled triangle is #pi/6#
By definition #sin pi/6 = 1/2# as that is the length (#1#) of the opposite side divide by the length (#2#) of the hypotenuse.
Now #sin (-alpha) = -sin alpha#, so #theta = -pi/6# is a solution of #sin theta = -1/2#.
Also #sin(pi-alpha) = sin(alpha)#
So #theta = pi - -pi/6 = (7pi)/6# is another solution.
#sin(alpha+2npi) = sin alpha# for all #n in ZZ#
So the solutions we have found are:
#theta = -pi/6 + 2npi# for all #n in ZZ#
#theta = (7pi)/6 + 2npi# for all #n in ZZ#
