<s>[INST] Find all values of $k$ in Trigonometry equation. Let $ \frac{\pi}{6} \le x \le \frac{\pi}{4} $  and $ \cot (x)= k  \sin(x) $ .

Find all values of $k$ such that satisfy in the above conditions.

We have $ k= \frac{\cos(x) }{\sin^2(x)} 
 =\frac{\cos(x) }{1-\cos^2(x)}$ now let $ f(x) =\frac{\cos(x) }{1-\cos^2(x)}$ then $ f^{'}$ is positive or negative on $ \frac{\pi}{6} \le x \le \frac{\pi}{4} $ ?
 [/INST] Taking $f(x) = \frac{\cos x}{\sin^2 x}$, we get $$f'(x) = -\csc x\left(\cot^2 x + \csc^2 x \right)$$
As $\cot^2 x + \csc^2 x > 0$ on $\left[\frac{\pi}{6}, \frac{\pi}{4}\right]$, the sign of $f'(x)$ is determined by that of $-\csc x$, which is negative on this interval. Therefore, $f'(x) < 0$ on $\left[\frac{\pi}{6}, \frac{\pi}{4}\right]$.
$k$ can take exactly the values in the range of $f$, on $\left[\frac{\pi}{6}, \frac{\pi}{4}\right]$. To find the range, we need only compute $f\left(\frac{\pi}{6}\right)$ and $f\left(\frac{\pi}{4}\right)$, since $f$ is strictly decreasing on this interval (so it takes all values between $f\left(\frac{\pi}{6}\right)$ and $f\left(\frac{\pi}{4}\right)$).
In particular, $f\left(\frac{\pi}{6}\right) = 2\sqrt 3$ and $f\left(\frac{\pi}{4}\right) = \sqrt 2$. Thus, $k$ takes values in the closed interval
$$[\sqrt 2, 2\sqrt 3]$$
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