We have:
#cos^-1(-1/2)=x#
Rearrange:
#-1/2=cos(x)#
Now we ask, which values of #x# would produce #-1/2# by the cosine function? This is possible because #-1/2# is a value we encounter regularly with the cosine and sine functions. We know that a right triangle with angle #60^o# adjacent to the horizontal side will give us a cosine of #1/2#. In the correct quadrant, namely quadrant #II#, we can get #-1/2#.
Where #a=1#.
The cosine in this case is given by #x/r# or the ratio of the adjacent side over the hypotenuse. This gives #cos(60^o)=1/2#.
In the second quadrant, #x<0# and #y>0#, giving #cos(120^o)=-1/2# (we use #120^o# because it gives us an angle of #60^o# above the negative #x#-axis).
Quadrant II:
The radian equivalent of #120^o# is #(2pi)/3#. Because the cosine function repeats every #2pi# units (i.e. has period of #2pi#), you may also write #arccos(-1/2)=(2pi)/3+-2pi#.
