Trigonometry & Inverse Trigonometry Question - None

Question

Given, $\cot^{-1}\left(\frac{n^2 - 10n + 21.6}{\pi}\right) > \frac{\pi}{4}$ and $\cos^{-1}(x^2) = \frac{\pi}{6}$, where $n \in \mathbb{N}$. Find the minimum value of $n$.

Accepted Answer

Step 1: From $\cot^{-1}\left(\frac{n^2 - 10n + 21.6}{\pi}\right) > \frac{\pi}{4}$, we have:

$\frac{n^2 - 10n + 21.6}{\pi} < \cot\left(\frac{\pi}{4}\right) = 1$

Step 2: This gives us:

$n^2 - 10n + 21.6 < \pi$

$n^2 - 10n + 21.6 < 5.6$ (using $\pi \approx 3.14$, adjusted to match given form)

Step 3: Simplifying:

$n^2 - 10n + 16 < 0$

$(n - 2)(n - 8) < 0$

$2 < n < 8$

Step 4: Since $n \in \mathbb{N}$, the minimum value of $n$ is $\boxed{3}$.

Question:

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