$ \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?
$ \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?
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 $\sin 2x = 0$.