I have the equation $ \ \sin(2x-1) \ = \ 0 $ . I know the answer is $ \ .5 + \pi k \ $ . but I don't understand how the answer was derived if the domain for sine must be between ($ -\pi/2 , \pi/2 ) $.
I also have $ \sin^2(x-\pi) \ + \ 2 \sin(x-\pi) \ = \ 0 $ ; I know you have to pull out $ \sin(x- \pi) \ $ to get:
$ \ \sin(x-\pi) [\sin(x-\pi)+2] \ $ but I do not know how to go from there.