Solve the following equations for 0≤0≤360°: i) 2 Sec20+tan 0−3=0 ii) Cos20 + Sin 0+1=0

Factor the quadratic equation:

This gives two possible values for $y$:

So, $\tan \theta = \frac{1}{2}$ or $\tan \theta = -1$.

Since $\tan \theta$ is positive, $\theta$ is in Quadrant I or Quadrant III.
The reference angle is $\alpha = \arctan\left(\frac{1}{2}\right) \approx 26.565^\circ$.
In Quadrant I: $\theta_1 = \alpha \approx 26.57^\circ$.
In Quadrant III: $\theta_2 = 180^\circ + \alpha \approx 180^\circ + 26.565^\circ = 206.565^\circ \approx 206.57^\circ$.

Since $\tan \theta$ is negative, $\theta$ is in Quadrant II or Quadrant IV.
The reference angle is $\alpha = \arctan(1) = 45^\circ$.
In Quadrant II: $\theta_3 = 180^\circ - \alpha = 180^\circ - 45^\circ = 135^\circ$.
In Quadrant IV: $\theta_4 = 360^\circ - \alpha = 360^\circ - 45^\circ = 315^\circ$.
The solutions are $\boxed{26.57^\circ, 135^\circ, 206.57^\circ, 315^\circ}$.

Multiply by $-1$ to rearrange into standard quadratic form:

Factor the quadratic equation:

This gives two possible values for $y$:

So, $\sin \theta = 2$ or $\sin \theta = -1$.

For $\sin \theta = 2$: There is no solution because the range of the sine function is $[-1, 1]$.
For $\sin \theta = -1$: This occurs at $\theta = 270^\circ$.
The solution is $\boxed{270^\circ}$.