The problem presents a trigonometric equation that can be solved by treating it as a quadratic equation. We will solve the equation:

The problem presents a trigonometric equation that can be solved by treating it as a quadratic equation. We will solve the equation:

$$2 \sin^2 \theta - \sin \theta - 1 = 0$$

Let $x = \sin \theta$. Substituting this into the equation gives a standard quadratic form:

$$2x^2 - x - 1 = 0$$

We can factor the quadratic equation. We look for two numbers that multiply to $(2)(-1) = -2$ and add up to $-1$. These numbers are $-2$ and 1.

$$2x^2 - 2x + x - 1 = 0$$

Factor by grouping:

$$2x(x - 1) + 1(x - 1) = 0$$ $$(2x + 1)(x - 1) = 0$$

This yields two possible values for $x$:

$$2x + 1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2}$$ $$x - 1 = 0 \implies x = 1$$

Now we substitute $\sin \theta$ back for $x$:

Case 1: $\sin \theta = 1$
The general solution for $\sin \theta = 1$ is when $\theta$ is $\frac{\pi}{2}$ plus any multiple of $2\pi$.

$$\theta = \frac{\pi}{2} + 2n\pi$$

where $n$ is an integer.

Case 2: $\sin \theta = -\frac{1}{2}$
The reference angle for $\sin \alpha = \frac{1}{2}$ is $\alpha = \frac{\pi}{6}$ ($30^\circ$). Since $\sin \theta$ is negative, $\theta$ must be in the third or fourth quadrants.
In the third quadrant:

$$\theta = \pi + \frac{\pi}{6} = \frac{6\pi}{6} + \frac{\pi}{6} = \frac{7\pi}{6}$$

In the fourth quadrant:

$$\theta = 2\pi - \frac{\pi}{6} = \frac{12\pi}{6} - \frac{\pi}{6} = \frac{11\pi}{6}$$

The general solutions for $\sin \theta = -\frac{1}{2}$ are:

$$\theta = \frac{7\pi}{6} + 2n\pi$$ $$\theta = \frac{11\pi}{6} + 2n\pi$$

where $n$ is an integer.

Combining all possible values for $\theta$, the general solutions are:

$$\boxed{\theta = \frac{\pi}{2} + 2n\pi, \quad \theta = \frac{7\pi}{6} + 2n\pi, \quad \theta = \frac{11\pi}{6} + 2n\pi \quad \text{where } n \in \mathbb{Z}}$$