In part (b), where you wrote the equation as

$y = \frac{1}{6} \cos \left( 2 \left( x + \frac{\pi}{3} \right) \right),$

it seems there is an issue with the frequency factor inside the cosine function. Since the period of a cosine function is defined as $\frac{2\pi}{k}$, and you've identified the period as $\pi$, the correct value of $k$ should be $k = \frac{2\pi}{\pi} = 2$.

This part of your equation is correct, as you used the factor 2. However, the correct form of the equation should account for the horizontal shift more carefully. Given that the horizontal shift is $-\frac{\pi}{3}$, we should replace $x + \frac{\pi}{3}$ with $x - \left(-\frac{\pi}{3}\right) = x + \frac{\pi}{3}$.

Therefore, the corrected equation for the cosine function in the form $y = a \cos(k(x - b))$ should be:

$y = \frac{1}{6} \cos \left( 2 \left( x + \frac{\pi}{3} \right) \right)$

If you tried this form and it didn't work, double-check the signs and ensure the angle inside the cosine function is correctly entered.

Let me know if you have any questions or need further clarification on this!

Here are five related questions you might find helpful:

1. How is the period of a trigonometric function affected by the coefficient inside the function?
2. What is the relationship between phase shift and horizontal shift in trigonometric functions?
3. How would the equation change if the amplitude or phase shift was different?
4. Why is the period of cosine functions $\frac{2\pi}{k}$?
5. Can you explain how to determine the horizontal shift from a graph?

Tip: For trigonometric transformations, always identify amplitude, period, and phase shift separately to construct the correct equation.