We are tasked with finding the equation of the tangent line to the curve $y = 5 \sin^2(x)$ at $x = \frac{\pi}{6}$. To do this, we need to:

1. Differentiate $y$ with respect to $x$ to find the slope of the tangent line at $x = \frac{\pi}{6}$.
2. Use the point-slope form of the tangent line equation to find the equation of the tangent.

Step 1: Differentiate $y = 5 \sin^2(x)$

First, rewrite the equation using a chain rule-friendly form: $y = 5 (\sin(x))^2$ Now apply the chain rule to differentiate: $\frac{dy}{dx} = 5 \cdot 2 \sin(x) \cdot \cos(x) = 10 \sin(x) \cos(x)$ Using the double-angle identity $\sin(2x) = 2 \sin(x) \cos(x)$, we simplify the derivative: $\frac{dy}{dx} = 5 \sin(2x)$

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

Substitute $x = \frac{\pi}{6}$ into the derivative to find the slope of the tangent line: $\frac{dy}{dx}\bigg|_{x = \frac{\pi}{6}} = 5 \sin\left(2 \cdot \frac{\pi}{6}\right) = 5 \sin\left(\frac{\pi}{3}\right) = 5 \cdot \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2}$ Thus, the slope of the tangent line is $\frac{5\sqrt{3}}{2}$.

Step 3: Find the point on the curve at $x = \frac{\pi}{6}$

We now find the $y$-coordinate of the point where $x = \frac{\pi}{6}$: $y = 5 \sin^2\left(\frac{\pi}{6}\right) = 5 \left(\frac{1}{2}\right)^2 = 5 \cdot \frac{1}{4} = \frac{5}{4}$ So the point on the curve is $\left(\frac{\pi}{6}, \frac{5}{4}\right)$.

Step 4: Use the point-slope form to find the equation of the tangent

The point-slope form of a line is: $y - y_1 = m(x - x_1)$ where $m$ is the slope, and $(x_1, y_1)$ is the point on the curve.

Substitute the slope $\frac{5\sqrt{3}}{2}$ and the point $\left(\frac{\pi}{6}, \frac{5}{4}\right)$ into the formula: $y - \frac{5}{4} = \frac{5\sqrt{3}}{2}\left(x - \frac{\pi}{6}\right)$ This is the equation of the tangent line.

Would you like me to simplify this further or provide additional details?

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Here are some related questions:

1. How do we use the product rule for differentiating trigonometric functions?
2. What is the geometric meaning of a tangent line to a curve?
3. How would the tangent equation change if $x = \frac{\pi}{3}$ instead?
4. What is the significance of the double-angle identity in simplifying derivatives?
5. How do we find the normal line to a curve at a given point?

Tip: When differentiating trigonometric functions, identities like the double-angle identity can often simplify your work.