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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, $$2x-1=n\pi\\ \implies \boxed{x=\dfrac{n\pi+1}{2}=0.5+k\pi,\quad k=\dfrac{n}{2}\forall n\in\mathbb{Z}}$$

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