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$ \sin(2x) \cdot \cos(2x) + \sin(2x) = 0 $

In the correction model I have something I don't understand is done in the first step:

$ \sin 2x(\cos 2x + 1) = 0 $

Is this step correct? And can someone explain that step to me?

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That step is correct, and is performed by taking $\sin 2x$ out as a common factor. In general, $$ab + a = a(b+1)$$ for any numbers $a,b$.

The usual error made in these questions is to act upon the temptation to divide by $\sin 2x$, in which case you obtain all the solutions corresponding to $\cos 2x = -1$, but miss out all the solutions corresponding to $\sin 2x = 0$.

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Thanks! I thought about it but for some reason I didn't understand it.. thanks! – ylykcoitus Sep 15 '12 at 11:10

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