To solve the equation \cos(x)(2\sin(x)+1) = 0 , we can set each factor equal to zero:
1. Set \cos(x) = 0 :
\cos(x) = 0
x = \frac{\pi}{2} + k\pi where k is an integer.
2. Set 2\sin(x) + 1 = 0 :
2\sin(x) + 1 = 0
2\sin(x) = -1
\sin(x) = -\frac{1}{2}
x = \frac{7\pi}{6} + 2k\pi, \ \frac{11\pi}{6} + 2k\pi where k is an integer.
\boxed{x = \frac{\pi}{2} + k\pi, \ \frac{7\pi}{6} + 2k\pi, \ \frac{11\pi}{6} + 2k\pi}