Task B-3.1. Determine the value of the expression $\frac{\sin x}{2-3 \cos x}$ if ctg $\frac{x}{2}=\frac{1}{2}$.
## First Solution.

$$
\begin{aligned}
\frac{\sin x}{2-3 \cos x} & =\frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2\left(\cos ^{2} \frac{x}{2}+\sin ^{2} \frac{x}{2}\right)-3\left(\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}\right)} \\
& =\frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{-\cos ^{2} \frac{x}{2}+5 \sin ^{2} \frac{x}{2}}
\end{aligned}
$$

If we divide the numerator and the denominator in the last equality by $\sin ^{2} \frac{x}{2}$, we get

$$
\frac{\sin x}{2-3 \cos x}=\frac{2 \operatorname{ctg} \frac{x}{2}}{-\operatorname{ctg}^{2} \frac{x}{2}+5}=\frac{1}{-\frac{1}{4}+5}=\frac{4}{19}
$$