It looks like you're dealing with a limit involving trigonometric functions as \( x \) approaches \( \frac{\pi}{6} \). Let me clarify the expression and solve it step by step.

We want to evaluate:

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

### Step 1: Evaluate \(\cos(3x)\) as \(x \to \frac{\pi}{6}\)

When \( x = \frac{\pi}{6} \):
\[
\cos(3x) = \cos\left( 3 \times \frac{\pi}{6} \right) = \cos\left( \frac{\pi}{2} \right) = 0
\]

### Step 2: Evaluate \(1 - 2\sin(x)\) as \(x \to \frac{\pi}{6}\)

When \( x = \frac{\pi}{6} \):
\[
\sin\left( \frac{\pi}{6} \right) = \frac{1}{2}
\]
Thus,
\[
1 - 2\sin\left( \frac{\pi}{6} \right) = 1 - 2 \times \frac{1}{2} = 1 - 1 = 0
\]

### Step 3: Apply L'Hopital's Rule

Since the limit results in the indeterminate form \( \frac{0}{0} \), we can apply **L'Hopital's Rule**, which states that:
\[
\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}
\]
if the limit on the right-hand side exists.

#### Differentiate the numerator \( \cos(3x) \):
\[
\frac{d}{dx}[\cos(3x)] = -3\sin(3x)
\]

#### Differentiate the denominator \( 1 - 2\sin(x) \):
\[
\frac{d}{dx}[1 - 2\sin(x)] = -2\cos(x)
\]

Thus, we now need to evaluate the new limit:
\[
\lim_{x \to \frac{\pi}{6}} \frac{-3\sin(3x)}{-2\cos(x)} = \lim_{x \to \frac{\pi}{6}} \frac{3\sin(3x)}{2\cos(x)}
\]

### Step 4: Evaluate the new limit

When \( x = \frac{\pi}{6} \), we have:
\[
\sin\left( 3 \times \frac{\pi}{6} \right) = \sin\left( \frac{\pi}{2} \right) = 1
\]
and
\[
\cos\left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}
\]

So, the limit becomes:
\[
\frac{3 \times 1}{2 \times \frac{\sqrt{3}}{2}} = \frac{3}{\sqrt{3}} = \sqrt{3}
\]

### Final Answer:

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

Would you like further details, or do you have any questions?

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Here are 5 related questions for further exploration:

1. How do you apply L'Hopital's Rule for higher-order derivatives?
2. What is the geometric meaning of \(\sin(x)\) and \(\cos(x)\) near small angles?
3. How can limits involving trigonometric identities be simplified without L'Hopital's Rule?
4. How does the derivative of \(\cos(3x)\) relate to chain rule applications?
5. What other indeterminate forms can occur in limits, besides \(\frac{0}{0}\)?

**Tip:** Always check if L'Hopital's Rule can be applied by confirming the \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) form before differentiating!

As x → π/6, evaluate the limit of cos(3x) / (1 − 2 sin(x))

Learn how to solve the trigonometric limit of cos(3x) divided by (1 − 2sin(x)) as x approaches π/6 using L'Hopital's Rule. This detailed step-by-step solution covers the differentiation of trigonometric functions and application of limits, ideal for calculus students.

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