What's wrong with this transformation?

There is no problem. Your solution is correct, and the other solution is correct. But you need to do a little work to see that they give the same answers:

• $\cos \frac {3x}2 = 0$ gives $x = \frac{(2k+1)\pi}3:\ x \in \{\pm \frac{\pi}3, \pm\pi,\pm \frac{5\pi}3, \pm \frac{7\pi}3, \pm 3\pi, \ldots\}$
• $\sin \frac {x}2 = 0$ gives $x = 2k\pi:\ x \in \{0, \pm 2\pi, \pm 4\pi, \ldots\}$

while

• $\cos x = \frac 12$ gives $x =\pm \frac{\pi}3 + 2k\pi:\ x \in \{\pm \frac{\pi}3, \pm \frac{5\pi}3, \pm \frac{7\pi}3, \ldots\}$
• $\sin x = 0$ gives $x = k\pi:\ x \in \{0, \pm\pi, \pm 2\pi, \pm 3\pi,\pm 4\pi, \ldots\}$

Comparison of the values shows that the same ones are in both lists.

You must disregard all points where $\cos(x) = 0$. These are outside of the domain of the initial expression you have on the left-hand side.

The solutions to $\sin(\frac{x}{2})=0$ or $\cos(\frac{3x}{2})=0$ are given by $$\frac{x}{2} = \pi k \text{ or } \frac{3x}{2} = \frac{\pi}{2}+\pi k, k \in \mathbb Z,$$ which can be rewritten as $$x = 2\pi k \text{ or } x = \frac{\pi}{3}+\frac{2\pi}{3}k, k \in \mathbb Z.$$

The solutions to $\cos(x) = \frac{1}{2}$ or $\sin(x)=0$ are given by $$x = \frac{\pi}{3}+2\pi k, \frac{5\pi}{3}+2\pi k, \pi k, k \in \mathbb Z.$$ These are the same sets of points.