1. The problem asks us to find all solutions to the equation $$\sin x = -1$$ in the interval $$-180^\circ \leq x < 90^\circ$$. 2. Recall the general solution for $$\sin x = -1$$ is given by $$x = -90^\circ + 360^\circ k$$, where $$k$$ is any integer. 3. To find solutions in the given interval, we plug in integer values of $$k$$ and check if the resulting $$x$$ lies within $$-180^\circ \leq x < 90^\circ$$. 4. For $$k=0$$: $$x = -90^\circ + 360^\circ \times 0 = -90^\circ$$, which is within the interval. 5. For $$k=1$$: $$x = -90^\circ + 360^\circ = 270^\circ$$, which is outside the interval. 6. For $$k=-1$$: $$x = -90^\circ - 360^\circ = -450^\circ$$, which is less than $$-180^\circ$$, so outside the interval. 7. Therefore, the only solution in the interval is $$x = -90^\circ$$. 8. Also, the given notes state the vertical asymptote is at $$x=2$$ and the horizontal asymptote is at $$y=t$$ (value unknown). 9. From the initial info, the angle $$\theta$$ with $$\sin \theta = 1$$ in the interval $$0^\circ \leq \theta \leq 90^\circ$$ is $$\theta = 90^\circ$$. **Final answers:** - Solutions to $$\sin x = -1$$ in $$-180^\circ \leq x < 90^\circ$$: $$x = -90^\circ$$ - Vertical asymptote: $$x = 2$$ - Horizontal asymptote: $$y = t$$ (value not given) - Angle $$\theta = 90^\circ$$ where $$\sin \theta = 1$$