Subtract 1, divide by 2.
#sinx = -1/2#
#sinx = pm1/2# at #30^o#, #150^o#, #210^o#, and #330^o#.
#sinx = -1/2# at #210^o# and #330^o#.
So, you have two answers. Really though, you have infinite answers if you consider that there are equivalent angles at every #pm360^o#. What you get, then, is:
#210^o/180^o*pi = (7pi)/6#
#330^o/180^o*pi = (11pi)/6#
Therefore, #sinx = -1/2# at #(7pi)/6#, #(11pi)/6#, #(19pi)/6#, #(23pi)/6#, etc.
Notice how #(11pi)/6# = #2pi - pi/6#, and #(7pi)/6 = 0 + (7npi)/6#. Recall how #sin 0 = sin 2pi = sin 2npi#, where n#in ZZ# (n is in the set of integers). Thus, you can write:
#x = 2npi-pi/6#
#x = 2npi+(7pi)/6#
