8.009. $\operatorname{ctg} t-\sin t=2 \sin ^{2} \frac{t}{2}$.
## Solution.

Domain of definition: $\sin t \neq 0$.

$$
\begin{aligned}
& \frac{\cos t}{\sin t}-\sin t=1-\cos t \Rightarrow \cos t-\sin ^{2} t=\sin t-\sin t \cos t \\
& (\cos t+\sin t \cos t)-\left(\sin ^{2} t+\sin t\right)=0, \cos t(1+\sin t)-\sin t(1+\sin t)=0 \\
& (1+\sin t)(\cos t-\sin t)=0
\end{aligned}
$$

From this, 1) $1+\sin t=0$ or 2) $\cos t-\sin t=0$. Then:

1) $\sin t=-1, t_{1}=-\frac{\pi}{2}+2 \pi k=\frac{\pi}{2}(4 k-1), k \in Z$;
2) $\cos t=\sin t \Leftrightarrow \operatorname{tg} t=1, t_{2}=\frac{\pi}{4}+\pi n=\frac{\pi}{4}(4 n+1), \quad n \in Z$.

Answer: $\quad t_{1}=\frac{\pi}{2}(4 k-1) ; t_{2}=\frac{\pi}{4}(4 n+1) \cdot k, n \in Z$.