# Solving for a Sine equation

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.

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The domain for $\sin$ can be anything. It's just that all inputs can be reduced to just an input between $-\pi$ and $\pi$, because $\sin(\theta) = \sin(\theta+2\pi) = \sin(\theta+4\pi)$ and so on. –  user137794 Apr 14 '14 at 22:06

The domain of $\sin$ is $(-\infty,\infty)$. Now, $\sin \theta=0$ if $\theta=n\pi,\space n\in\mathbb{Z}$. Thus, $$2x-1=n\pi\\ \implies \boxed{x=\dfrac{n\pi+1}{2}=0.5+k\pi,\quad k=\dfrac{n}{2}\forall n\in\mathbb{Z}}$$