To solve the following limit:

\[
\lim_{{x \to \frac{\pi}{6}}} \frac{\cos(3x)}{1 - 2 \sin(x)}
\]

### Step-by-Step Solution:

1. **Substitute \( x = \frac{\pi}{6} \) into the expression:**

\[
\cos(3x) = \cos\left(3 \cdot \frac{\pi}{6}\right) = \cos\left(\frac{\pi}{2}\right)
\]

Since \( \cos\left(\frac{\pi}{2}\right) = 0 \), we have:

\[
\cos(3x) = 0
\]

2. **Now evaluate the denominator:**

\[
\sin(x) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}
\]

Thus, the denominator becomes:

\[
1 - 2 \sin\left(\frac{\pi}{6}\right) = 1 - 2 \times \frac{1}{2} = 1 - 1 = 0
\]

3. **Form of the Limit:**

Now the limit is in the indeterminate form \( \frac{0}{0} \), so we can apply **L'Hôpital's Rule**. To apply L'Hôpital's Rule, we differentiate the numerator and the denominator with respect to \(x\).

4. **Differentiate the numerator:**

\[
\frac{d}{dx} \left[ \cos(3x) \right] = -3 \sin(3x)
\]

5. **Differentiate the denominator:**

\[
\frac{d}{dx} \left[ 1 - 2 \sin(x) \right] = -2 \cos(x)
\]

6. **Apply L'Hôpital's Rule:**

The limit now becomes:

\[
\lim_{{x \to \frac{\pi}{6}}} \frac{-3 \sin(3x)}{-2 \cos(x)}
\]

7. **Substitute \( x = \frac{\pi}{6} \) into the new expression:**

\[
\sin(3x) = \sin\left( \frac{\pi}{2} \right) = 1
\]
\[
\cos\left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}
\]

Now the limit becomes:

\[
\frac{-3 \times 1}{-2 \times \frac{\sqrt{3}}{2}} = \frac{-3}{-\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}
\]

### Final Answer:

\[
\lim_{{x \to \frac{\pi}{6}}} \frac{\cos(3x)}{1 - 2 \sin(x)} = \sqrt{3}
\]

Would you like further details or clarifications?

### Relative Questions:
1. How does L'Hôpital's Rule help solve indeterminate forms?
2. What are other methods to solve limits besides L'Hôpital's Rule?
3. How can you check if a limit leads to an indeterminate form?
4. What are common trigonometric limits and their applications?
5. How does the behavior of sine and cosine affect limits?

### Tip:
When you encounter \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) forms, always consider using L'Hôpital's Rule!

Explore how to solve the limit lim x→π/6 [cos(3x) / (1 − 2sin(x))] using L'Hôpital's Rule and trigonometric identities. This detailed solution breaks down each step, from substitution to applying the rule, making it ideal for students in Grades 11-12 or early college.

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