## Task A-3.6. (10 points)

Determine all values of the real parameter $a$ for which the equation

$$
8^{a \sin ^{2} x}=4 \cdot 2^{\cos ^{2} x}
$$

has exactly one solution in the interval $\left[-\frac{\pi}{6}, \frac{2 \pi}{3}\right]$.
## Solution.

The given equation can be written in the form $2^{3 a \sin ^{2} x}=2^{2+\cos ^{2} x}$, from which it follows that $3 a \sin ^{2} x=2+\cos ^{2} x$.

Now we have $3 a \sin ^{2} x=3-\sin ^{2} x$, or $\sin ^{2} x=\frac{3}{3 a+1}$.

Since $\sin ^{2}(-x)=\sin ^{2} x, \quad\left|\sin \left(-\frac{\pi}{6}\right)\right|=\left|\sin \frac{\pi}{6}\right|=\frac{1}{2} \quad$ and $\quad \sin \frac{2 \pi}{3}=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}$, by examining the graph

![](https://cdn.mathpix.com/cropped/2024_05_30_13589bc3b5dd31895527g-17.jpg?height=679&width=1405&top_left_y=1111&top_left_x=263)

we conclude that the solution to the given equation in the interval $\left[-\frac{\pi}{6}, \frac{2 \pi}{3}\right]$ will be unique if and only if $\sin x$ is in the set $\left\langle\frac{1}{2}, \frac{\sqrt{3}}{2}\right\rangle \cup\{1\}$.

Now we can easily conclude that it must be $\frac{1}{4}<\frac{3}{3 a+1}<\frac{3}{4}$ or $\frac{3}{3 a+1}=1$, from which we get $1<a<\frac{11}{3}$ or $a=\frac{2}{3}$.

Note: A student who omits the solution $a=\frac{2}{3}$ can have a maximum of 9 points.