The given equation for the depth of water is:

$h(t) = 8 + 4 \cos\left(\frac{\pi t}{6}\right)$

We need to find for how long the water is less than 6 metres deep, i.e., when $h(t) < 6$.

Step 1: Set up the inequality

$8 + 4 \cos\left(\frac{\pi t}{6}\right) < 6$

Step 2: Simplify the inequality

$4 \cos\left(\frac{\pi t}{6}\right) < 6 - 8$ $4 \cos\left(\frac{\pi t}{6}\right) < -2$ $\cos\left(\frac{\pi t}{6}\right) < -\frac{1}{2}$

Step 3: Solve for $t$

The cosine function is less than $-\frac{1}{2}$ when the angle is in the second or third quadrant. Specifically, this occurs for:

$\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$

Thus, we need to solve for $\frac{\pi t}{6}$ in the intervals where the cosine is less than $-\frac{1}{2}$:

$\frac{\pi t}{6} \in \left[\frac{2\pi}{3}, \frac{4\pi}{3}\right] + 2k\pi$

For the principal solution (since we're looking at one day, $t \in [0, 24]$):

$\frac{2\pi}{3} \leq \frac{\pi t}{6} \leq \frac{4\pi}{3}$

Step 4: Solve for $t$

Multiply through by 6 to isolate $t$:

$2 \leq t \leq 4$

Thus, the water is less than 6 metres deep between 2 hours and 4 hours after midnight.

Step 5: Conclusion

The water is less than 6 metres deep for exactly 2 hours on this day, which is equivalent to 120 minutes.

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Would you like more details or have any questions about the process?

Here are 5 related questions for further exploration:

1. How would the time interval change if the cosine term's coefficient was different?
2. What is the maximum depth of the water according to the given function?
3. At what time is the depth exactly 8 metres?
4. Can you explain how to graph the function $h(t) = 8 + 4 \cos\left(\frac{\pi t}{6}\right)$?
5. How would the answer change if the period of the cosine function was modified?

Tip: Always remember that cosine functions have symmetrical properties, making it easier to solve trigonometric inequalities.