Save for theta given that sin2 theta - 3/2 sin theta − 1 = 0 for 0 degrees <= theta <= 360 degrees. Determine the average rate of cooling between second and tenth minutes.

Substitute this into the given equation.

This gives two possible values for $k$:

The range of the sine function is $[-1, 1]$. Therefore, $\sin \theta = 2$ has no solution.
For $\sin \theta = -\frac{1}{2}$:
The reference angle for $\sin \alpha = \frac{1}{2}$ is $\alpha = 30^\circ$.
Since $\sin \theta$ is negative, $\theta$ lies in the third and fourth quadrants.
In the third quadrant: $\theta = 180^\circ + 30^\circ = 210^\circ$.
In the fourth quadrant: $\theta = 360^\circ - 30^\circ = 330^\circ$.
Both angles are within the given range $0^\circ \le \theta < 360^\circ$.

The solutions for $\theta$ are $\boxed{210^\circ \text{ or } 330^\circ}$.

At the second minute ($t_1 = 2 \text{ min}$), the temperature is $T_1 = 70^\circ \text{C}$.
At the tenth minute ($t_2 = 10 \text{ min}$), the temperature is $T_2 = 20^\circ \text{C}$.

Change in temperature ($\Delta T$) = $T_2 - T_1 = 20^\circ \text{C} - 70^\circ \text{C} = -50^\circ \text{C}$.
Change in time ($\Delta t$) = $t_2 - t_1 = 10 \text{ min} - 2 \text{ min} = 8 \text{ min}$.

Rate $= \frac{\Delta T}{\Delta t}$.

The average rate of cooling is $\boxed{-6.25 \frac{^\circ \text{C}}{\text{min}}}$.