To find the equation of the tangent line to the curve $y = 6 \sin(x)$ at the point $\left( \frac{\pi}{6}, 3 \right)$, we follow these steps:

Step 1: Differentiate the function to find the slope of the tangent line.

The function is $y = 6 \sin(x)$. The derivative $y'$ gives the slope of the tangent line at any point $x$. $\frac{dy}{dx} = \frac{d}{dx} \left( 6 \sin(x) \right) = 6 \cos(x)$

Step 2: Evaluate the derivative at $x = \frac{\pi}{6}$.

To find the slope at $\left( \frac{\pi}{6}, 3 \right)$, substitute $x = \frac{\pi}{6}$ into the derivative: $\text{slope} = 6 \cos\left( \frac{\pi}{6} \right)$ Since $\cos\left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}$, the slope becomes: $\text{slope} = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3}$ Thus, the slope $m = 3\sqrt{3}$.

Step 3: Use the point-slope form to find the equation of the tangent line.

The point-slope form of a line is: $y - y_1 = m(x - x_1)$ We know $m = 3\sqrt{3}$ and the point $\left( x_1, y_1 \right) = \left( \frac{\pi}{6}, 3 \right)$. Substituting these values: $y - 3 = 3\sqrt{3}\left(x - \frac{\pi}{6}\right)$

Step 4: Solve for $y$ in the slope-intercept form $y = mx + b$.

Expand the equation: $y - 3 = 3\sqrt{3}x - 3\sqrt{3}\cdot\frac{\pi}{6}$ $y - 3 = 3\sqrt{3}x - \frac{\pi\sqrt{3}}{2}$ Now, add 3 to both sides to isolate $y$: $y = 3\sqrt{3}x - \frac{\pi\sqrt{3}}{2} + 3$

Thus, the equation of the tangent line is: $y = 3\sqrt{3}x + \left( 3 - \frac{\pi\sqrt{3}}{2} \right)$

So, $m = 3\sqrt{3}$ and $b = 3 - \frac{\pi\sqrt{3}}{2}$.

Would you like more details on any of the steps?

Here are five related questions to further your understanding:

1. What is the geometric meaning of the slope of the tangent line?
2. How would the equation change if the function were $y = 6 \cos(x)$ instead?
3. What happens to the tangent line equation if we change the point of tangency?
4. Can we generalize this process to any sinusoidal function, such as $y = A \sin(x)$?
5. How does the value of $b$ change if the point of tangency is altered?

Tip: In calculus, the derivative of a function at a point gives the slope of the tangent line at that specific point, which helps in constructing the tangent equation.