Solve for theta in the interval 0 <= theta <= 2pi: 2 theta theta = 1

The general solution for $\sin x = 1$ is $x = \frac{\pi}{2} + 2n\pi$, where $n$ is an integer.
So, $2\theta = \frac{\pi}{2} + 2n\pi$.

For $n=0$: $\theta = \frac{\pi}{4}$
For $n=1$: $\theta = \frac{\pi}{4} + \pi = \frac{5\pi}{4}$
For $n=2$: $\theta = \frac{\pi}{4} + 2\pi = \frac{9\pi}{4}$, which is outside the interval.

The solutions are $\frac{\pi}{4}$ and $\frac{5\pi}{4}$.

12. Solve for $\theta$ in the interval $0^\circ \leq \theta \leq 360^\circ$:

Case 1: $\cos \theta = 0$
For $0^\circ \leq \theta \leq 360^\circ$, $\cos \theta = 0$ when $\theta = 90^\circ$ or $\theta = 270^\circ$.

Case 2: $2 \sin \theta - 1 = 0$

For $0^\circ \leq \theta \leq 360^\circ$, $\sin \theta = \frac{1}{2}$ when $\theta = 30^\circ$ (in Quadrant I) or $\theta = 180^\circ - 30^\circ = 150^\circ$ (in Quadrant II).

The solutions are $30^\circ, 90^\circ, 150^\circ, 270^\circ$.

13. Solve for $\theta$ in the interval $0 \leq \theta \leq 2\pi$:

Multiply by $-1$ to make the leading coefficient positive.

Factor the quadratic equation:

Case 1: $2 \sin \theta + 1 = 0$

For $0 \leq \theta \leq 2\pi$, $\sin \theta = -\frac{1}{2}$ in Quadrant III and Quadrant IV.
The reference angle is $\frac{\pi}{6}$.
In Quadrant III: $\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$
In Quadrant IV: $\theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$

Case 2: $\sin \theta - 1 = 0$

For $0 \leq \theta \leq 2\pi$, $\sin \theta = 1$ when $\theta = \frac{\pi}{2}$.

The solutions are $\frac{\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6}$.