## Task B-4.3.

The numbers $\sin x$ and $\sin 2x$ are the first two terms of a geometric sequence where all terms are different from 0. Determine all numbers $x \in (0, 2\pi)$ for which the third term of this sequence is equal to $\sin 4x$.
## Solution.

The quotient of the geometric sequence is $q=\frac{\sin 2 x}{\sin x}=\frac{2 \sin x \cos x}{\sin x}=2 \cos x$, 1 point

From the condition $q \neq 0$, i.e., $\sin x \neq 0$ and $\sin 2 x \neq 0$, it follows that $x \neq \frac{k \pi}{2}, k \in \mathbb{Z}$.

Let $a_{3}=\sin 4 x$. Then, due to $a_{3}=a_{2} \cdot q$, we have

$$
\sin 4 x=\sin 2 x \cdot 2 \cos x
$$

Solving this, we get:

$$
\begin{aligned}
& 2 \sin 2 x \cos 2 x=2 \sin 2 x \cos x \\
& 2 \sin 2 x(\cos 2 x-\cos x)=0 \\
& \sin 2 x=0 \quad \text { or } \quad \cos 2 x-\cos x=0
\end{aligned}
$$

Since $\sin 2 x \neq 0$ due to $(*)$, 1 point

it follows that $\cos 2 x=\cos x$, from which $2 x= \pm x+2 k \pi$.

The solutions to this equation are $x=2 k \pi$ (does not satisfy $\left.\left(^{*}\right)\right)$ and $x=\frac{2 k \pi}{3}$.

Since $x \in\langle 0,2 \pi\rangle$, for $k=1$ and $k=2$ we obtain the desired solutions.

The third term of the given sequence is $\sin 4 x$ for $x \in\left\{\frac{2 \pi}{3}, \frac{4 \pi}{3}\right\}$.