Question

(2 cos x+3)(2 sin x-1)=0

Answer 1

2sinx - 1 = 0
2sinx = 1
sin x = 1/2
x = pi/6

Answer 2

On the other side for 2cosx + 3 = 0
cosx = -3/2
\[x = 2k \pi \pm \arccos(-3/2)\]

Answer 3

Oh, and in the end you can combine the two results so that
\[x \in \left\{ \frac{ \pi }{ 6 }\ , 2k \pi \pm acrccos(-\frac{ 3 }{ 2 }), k \in \mathbb{Z} \right\}\]