To solve the equation

\[
(1 - 2\sin x - \cos 2x)(3\cot x + \sqrt{3}) = 0,
\]

we can split it into two separate equations and solve each one.

### 1. Solve \(1 - 2\sin x - \cos 2x = 0\)

Recall the double angle identity:

\[
\cos 2x = 1 - 2\sin^2 x.
\]

Replace \(\cos 2x\) in the equation:

\[
1 - 2\sin x - (1 - 2\sin^2 x) = 0.
\]

Simplifying this gives:

\[
1 - 2\sin x - 1 + 2\sin^2 x = 0,
\]
\[
2\sin^2 x - 2\sin x = 0.
\]

Factoring out \(2\sin x\):

\[
2\sin x(\sin x - 1) = 0.
\]

Setting each factor to zero gives us:

1. \(2\sin x = 0 \implies \sin x = 0\)
2. \(\sin x - 1 = 0 \implies \sin x = 1\)

The solutions for \(\sin x = 0\) are:

\[
x = n\pi, \quad n \in \mathbb{Z}.
\]

The solution for \(\sin x = 1\) is:

\[
x = \frac{\pi}{2} + 2k\pi, \quad k \in \mathbb{Z}.
\]

### 2. Solve \(3 \cot x + \sqrt{3} = 0\)

Rearranging gives:

\[
3 \cot x = -\sqrt{3} \implies \cot x = -\frac{\sqrt{3}}{3}.
\]

The angle where \(\cot x = -\frac{\sqrt{3}}{3}\) is:

\[
x = \frac{5\pi}{6} + n\pi, \quad n \in \mathbb{Z}.
\]

### Final Solutions

1. From \(1 - 2\sin x - \cos 2x = 0\):
\[
x = n\pi, \quad n \in \mathbb{Z}
\]
\[
x = \frac{\pi}{2} + 2k\pi, \quad k \in \mathbb{Z}
\]

2. From \(3 \cot x + \sqrt{3} = 0\):
\[
x = \frac{5\pi}{6} + n\pi, \quad n \in \mathbb{Z}.
\]

Thus, the solutions to the original equation are:

\[
x = n\pi, \quad n \in \mathbb{Z}, \quad x = \frac{\pi}{2} + 2k\pi, \quad k \in \mathbb{Z}, \quad x = \frac{5\pi}{6} + n\pi, \quad n \in \mathbb{Z}.
\]