i've been having problems with these...
sin x= -.5 ,0 < x <2pi
sinx =-1/2
x=11pi/6 +2kpi
Not quite. Try thinking of it this way: We get a reference angle of .
Now, sine is negative in two quadrants, III and IV. So we have solutions for :
and
So the solution is
and
(You really don't need to change the arbitrary integer from k to m, but I do it to remind myself that the two do not need to be the same.)
-Dan