8.089. $\cos \left(2 t-18^{\circ}\right) \operatorname{tg} 50^{\circ}+\sin \left(2 t-18^{\circ}\right)=\frac{1}{2 \cos 130^{\circ}}$.
## Solution.

We have

$$
\begin{aligned}
& \frac{\cos \left(2 t-18^{\circ}\right) \sin 50^{\circ}}{\cos 50^{\circ}}+\sin \left(2 t-18^{\circ}\right)=\frac{1}{2 \cos \left(180^{\circ}-50^{\circ}\right)} \Leftrightarrow \\
& \Leftrightarrow \frac{\cos \left(2 t-18^{\circ}\right) \sin 50^{\circ}+\sin \left(2 t-18^{\circ}\right) \cos 50^{\circ}}{\cos 50^{\circ}}=\frac{1}{-2 \cos 50^{\circ}} \Rightarrow \\
& \Rightarrow \sin \left(2 t-18^{\circ}+50^{\circ}\right)=-\frac{1}{2}, \quad \sin \left(2 t+32^{\circ}\right)=-\frac{1}{2}
\end{aligned}
$$

From this:

1) $2 t+32^{\circ}=-30^{\circ}+360^{\circ} k, \quad k \in \mathbb{Z}, t_{1}=-31^{\circ}+180^{\circ} k$;
2) $2 t+32^{\circ}=210^{\circ}+360^{\circ} k, t_{2}=89^{\circ}+180^{\circ} k, k \in \mathbb{Z}$.

Answer: $\quad t_{1}=-31^{\circ}+180^{\circ} k ; t_{2}=89^{\circ}+180^{\circ} k, k \in \mathbb{Z}$.