What's the difference between these two trig identity problems?

1) $2 \tan x=3 \sin x$
$2 \frac{\sin x}{\cos x}=3 \sin x \rightarrow$ $2 \sin x = 3 \sin x \cos x \rightarrow$ $\sin x(2 - 3 \cos x)=0$
a) $\sin x = 0 \rightarrow x =180n$
b) $\cos x = \frac{2}{3} \rightarrow x =$ 48.19 + 360n

2) $2 \sin 2x \cos ^2 x = \sin 2x$
$\sin 2x(2 \cos ^2 x -1) =0 \rightarrow$
a) $\sin 2x=0 \rightarrow 2x=180n \rightarrow x = 90n$
b) $\cos ^2 x = \frac{1}{2} \rightarrow \cos x = \pm \sqrt{\frac{1}{2} }$
$x = \pm$ 45 + 360n

of course the answer to number 2 is $\pm$ 45 + 180n
but the answer to number 1 is 48.19 + 360n! So what's the difference between the two? They are both cosine functions. I just thought always add 360 degrees because that's when the angle repeats itself.

(*fixed the bold...)