#### Answer

The two possible solutions are $\theta = 15^{\circ}$ and $\theta = 75^{\circ}$

#### Work Step by Step

$range = \frac{v_0^2~sin(2\theta)}{g}$
The maximum possible range occurs when $sin(2\theta) = 1$, that is, when $\theta = 45^{\circ}$. In that case, $range = \frac{v_0^2}{g}$
We need to find $\theta$ such that $range = \frac{1}{2}\times \frac{v_0^2}{g}$
$\frac{v_0^2~sin(2\theta)}{g} = \frac{1}{2}\times \frac{v_0^2}{g}$
$sin(2\theta) = 0.5$
$2\theta = arcsin(0.5)$
$2\theta = 30^{\circ}$
$\theta = 15^{\circ}$
Note that: $sin(180^{\circ}-2\theta) = sin(2\theta)$
We can find another possible solution.
$180^{\circ}-2\theta = arcsin(0.5)$
$180^{\circ}-2\theta = 30^{\circ}$
$\theta = 75^{\circ}$
The two possible solutions are $\theta = 15^{\circ}$ and $\theta = 75^{\circ}$.