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

1) $\displaystyle 2 \tan x=3 \sin x$

$\displaystyle 2 \frac{\sin x}{\cos x}=3 \sin x \rightarrow$ $\displaystyle 2 \sin x = 3 \sin x \cos x \rightarrow$ $\displaystyle \sin x(2 - 3 \cos x)=0$

a) $\displaystyle \sin x = 0 \rightarrow x =180n$

b) $\displaystyle \cos x = \frac{2}{3} \rightarrow x =$48.19 + **360n**

2) $\displaystyle 2 \sin 2x \cos ^2 x = \sin 2x$

$\displaystyle \sin 2x(2 \cos ^2 x -1) =0 \rightarrow$

a) $\displaystyle \sin 2x=0 \rightarrow 2x=180n \rightarrow x = 90n$

b) $\displaystyle \cos ^2 x = \frac{1}{2} \rightarrow \cos x = \pm \sqrt{\frac{1}{2} }$

$\displaystyle x = \pm$ 45 + **360n**

of course the answer to number 2 is $\displaystyle \pm$ 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...)