To find the equation of the tangent to the function $y = 2\sec(x)$ at $x = \frac{\pi}{6}$, follow these steps:

1. Differentiate the function:

The equation of the tangent line at a given point is found using the derivative of the function, which gives the slope of the tangent line.

We have $y = 2\sec(x)$.

The derivative of $\sec(x)$ with respect to $x$ is:

$\frac{d}{dx}[\sec(x)] = \sec(x)\tan(x)$

So, differentiating $y = 2\sec(x)$ with respect to $x$:

$\frac{dy}{dx} = 2\sec(x)\tan(x)$

This gives the slope of the tangent line at any point $x$.

2. Find the slope at $x = \frac{\pi}{6}$:

Now, substitute $x = \frac{\pi}{6}$ into the derivative to find the slope of the tangent line.

At $x = \frac{\pi}{6}$: $\sec\left(\frac{\pi}{6}\right) = \frac{1}{\cos\left(\frac{\pi}{6}\right)} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}$ $\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$

Thus, the slope at $x = \frac{\pi}{6}$ is:

$\frac{dy}{dx}\bigg|_{x = \frac{\pi}{6}} = 2 \cdot \frac{2}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}} = \frac{4}{3}$

3. Find the point of tangency:

At $x = \frac{\pi}{6}$, substitute into the original function to find the corresponding $y$-value.

$y = 2\sec\left(\frac{\pi}{6}\right) = 2 \cdot \frac{2}{\sqrt{3}} = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3}$

So, the point of tangency is $\left(\frac{\pi}{6}, \frac{4\sqrt{3}}{3}\right)$.

4. Write the equation of the tangent line:

The equation of a line is given by:

$y - y_1 = m(x - x_1)$

Where $m$ is the slope, and $(x_1, y_1)$ is the point of tangency.

Substituting the values $m = \frac{4}{3}$, $x_1 = \frac{\pi}{6}$, and $y_1 = \frac{4\sqrt{3}}{3}$:

$y - \frac{4\sqrt{3}}{3} = \frac{4}{3}\left(x - \frac{\pi}{6}\right)$

This is the equation of the tangent line.

Final Answer:

The equation of the tangent line to $y = 2\sec(x)$ at $x = \frac{\pi}{6}$ is:

$y - \frac{4\sqrt{3}}{3} = \frac{4}{3}\left(x - \frac{\pi}{6}\right)$

Let me know if you would like further details or have any questions.

Here are some related questions to deepen your understanding:

1. How do you find the equation of the tangent line for a general function at a specific point?
2. What is the difference between secant and tangent functions in terms of calculus and geometry?
3. Can you find the normal line at the same point $x = \frac{\pi}{6}$?
4. What is the significance of the derivative in determining the slope of a tangent?
5. How would the equation change if the function were $y = 3\sec(x)$?

Tip: Always check whether the function has vertical asymptotes near the point of tangency to ensure the tangent exists.