Find $24 \cot^2 x$ 
Suppose $x$ is in the interval $[0,\pi/2]$ and $\log_{24 \sin x} (24 \cos x) = \frac{3}{2}$. Find $24 \cot^2 x$.

From the given equation we have $24 \cos{x} = (24 \sin{x})^{\frac{3}{2}}$ and so $24\cot^2{x} = 24^2\sin{x}$. How do we continue?

A Solution
Since
$$
(24\cos(x))^2=(24\sin(x))^3\tag{1}
$$
we have
$$
24\sin^3(x)+\sin^2(x)-1=0\tag{2}
$$
We can try the rational root theorem to get
$$
\sin(x)=\frac13\tag{2}
$$
Now use
$$
24\cot^2(x)=24^2\sin(x)=192\tag{4}
$$

Uniqueness of the Solution
Since $\frac{\mathrm{d}}{\mathrm{d}u}\left(24u^3+u^2-1\right)=72u^2+2u$, the function $24u^3+u^2-1$ is increasing for $u\gt0$. This means there can only be one positive solution. Equation $(1)$ says that we need $\sin(x)\ge0$. Furthermore, the maximum of $24u^3+u^2-1$, for $u\le0$, is $-\frac{3887}{3888}$. Thus, there can be no negative solution.
