To solve the inequality \((1 - 2\sin x)(2\cos x + \sqrt{3}) < 0\), we'll analyze the product of the two factors separately. The product is negative when one factor is positive and the other is negative.

### Step 1: Analyze each factor

1. **Factor 1: \(1 - 2\sin x\)**

   \[
   1 - 2\sin x > 0 \quad \text{or} \quad 1 - 2\sin x < 0
   \]
   \[
   \sin x < \frac{1}{2} \quad \text{or} \quad \sin x > \frac{1}{2}
   \]

2. **Factor 2: \(2\cos x + \sqrt{3}\)**

   \[
   2\cos x + \sqrt{3} > 0 \quad \text{or} \quad 2\cos x + \sqrt{3} < 0
   \]
   \[
   \cos x > -\frac{\sqrt{3}}{2} \quad \text{or} \quad \cos x < -\frac{\sqrt{3}}{2}
   \]

### Step 2: Determine intervals where the product is negative

To determine where the product is negative, consider the cases:

1. **Case 1**: \(1 - 2\sin x > 0\) and \(2\cos x + \sqrt{3} < 0\)
   - \( \sin x < \frac{1}{2} \)
   - \( \cos x < -\frac{\sqrt{3}}{2} \)

2. **Case 2**: \(1 - 2\sin x < 0\) and \(2\cos x + \sqrt{3} > 0\)
   - \( \sin x > \frac{1}{2} \)
   - \( \cos x > -\frac{\sqrt{3}}{2} \)

### Step 3: Identify the solution intervals

1. **For Case 1**:
   - \( \sin x < \frac{1}{2} \) corresponds to \(x\) in \(\left[0, \frac{\pi}{6}\right) \cup \left(\frac{5\pi}{6}, 2\pi\right)\).
   - \( \cos x < -\frac{\sqrt{3}}{2} \) corresponds to \(x\) in \(\left(\frac{5\pi}{6}, \frac{7\pi}{6}\right)\).

   Intersection: \(\left(\frac{5\pi}{6}, \frac{7\pi}{6}\right)\).

2. **For Case 2**:
   - \( \sin x > \frac{1}{2} \) corresponds to \(x\) in \(\left(\frac{\pi}{6}, \frac{5\pi}{6}\right)\).
   - \( \cos x > -\frac{\sqrt{3}}{2} \) corresponds to \(x\) in \(\left[-\frac{\pi}{6}, \frac{5\pi}{6}\right)\).

   Intersection: \(\left(\frac{\pi}{6}, \frac{5\pi}{6}\right)\).

### Final Answer:
The solution to the inequality is the union of the intervals:

\[
\boxed{x \in \left(\frac{5\pi}{6}, \frac{7\pi}{6}\right) \cup \left(\frac{\pi}{6}, \frac{5\pi}{6}\right)}
\]

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**Relative Questions:**
1. How do we solve trigonometric inequalities involving multiple angles?
2. What is the graphical interpretation of the solution set on the unit circle?
3. How do we solve an inequality involving both \(\sin x\) and \(\cos x\)?
4. What are the specific values of \(\sin x\) and \(\cos x\) at key angles (like \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\))?
5. Can this method be applied to inequalities involving \(\tan x\) or \(\cot x\)?

**Tip:** When solving trigonometric inequalities, always consider the periodic nature of trigonometric functions, which might introduce multiple intervals for solutions.

Learn how to solve the trigonometric inequality (1 - 2sinx)(2cosx+rad3)<0 step-by-step with detailed analysis. This problem involves trigonometric identities and inequalities, focusing on the sine and cosine functions. Suitable for advanced high school students and learners interested in trigonometric inequalities.

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