## Task B-4.5.

Determine all natural numbers $y>1$ that satisfy the equation

$\log _{\sin x} y-3 \log _{y} \sqrt{\sin x}=\frac{1}{2}$, where $x \in\left[\frac{\pi}{6}, \frac{\pi}{3}\right]$.
## Solution.

Let's write the given equation in the following form:

$\log _{\sin x} y-\frac{3}{2} \cdot \frac{1}{\log _{\sin x} y}=\frac{1}{2}$.

Introduce the substitution $t=\log _{\sin x} y$ so that this equation becomes $t-\frac{3}{2} \cdot \frac{1}{t}=\frac{1}{2}$.

Note that $t=\log _{\sin x} y>0$ due to $y>1$, so the above equation is equivalent to $2 t^{2}-t-3=0$.

The solutions to this quadratic equation are $t_{1}=-1$ and $t_{2}=\frac{3}{2}$.

Thus, $\log _{\sin x} y=-1$ or $\log _{\sin x} y=\frac{3}{2}$, that is:

$y=\frac{1}{\sin x}$ or $y=(\sin x)^{\frac{3}{2}}$.

We discard the solution $y=(\sin x)^{\frac{3}{2}}$ because $y>1$, while $(\sin x)^{\frac{3}{2}} \leqslant 1$.

On the given interval $\left[\frac{\pi}{6}, \frac{\pi}{3}\right]$, we have $\frac{\pi}{6} \leqslant x \leqslant \frac{\pi}{3}$,

and since the sine function is increasing on this interval, it follows that:

$\sin \frac{\pi}{6} \leqslant \sin x \leqslant \sin \frac{\pi}{3}$, that is, $\frac{1}{2} \leqslant \sin x \leqslant \frac{\sqrt{3}}{2}$, so $\frac{2}{\sqrt{3}} \leqslant \frac{1}{\sin x} \leqslant 2$.

Thus, $y \in\left[\frac{2}{\sqrt{3}}, 2\right]$.

Since $y$ is a natural number and $y>1$, the only solution is $y=2$.