An angle $\theta$ with $0^{\circ} \leq \theta \leq 180^{\circ}$ satisfies $\sqrt{2} \cos 2 \theta=\cos \theta+\sin \theta$. Determine all possible values of $\theta$.
Using the identity $\cos 2 \theta=\cos ^{2} \theta-\sin ^{2} \theta$, we can rearrange $\sqrt{2} \cos 2 \theta=\cos \theta+\sin \theta$ to an equivalent equation.

$$
\begin{aligned}
\sqrt{2} \cos 2 \theta & =\cos \theta+\sin \theta \\
\sqrt{2}\left(\cos ^{2} \theta-\sin ^{2} \theta\right) & =\cos \theta+\sin \theta \\
\sqrt{2}(\cos \theta+\sin \theta)(\cos \theta-\sin \theta) & =\cos \theta+\sin \theta \\
(\cos \theta+\sin \theta)(\sqrt{2}(\cos \theta-\sin \theta)-1) & =0 .
\end{aligned}
$$

It follows that $\theta$ satisfies the equation if and only if either $\cos \theta+\sin \theta=0$ or $\cos \theta-\sin \theta=\frac{1}{\sqrt{2}}$. $\cos \theta+\sin \theta=0$ implies $\tan \theta=-1$, and the only angle $\theta$ satisfying $\tan \theta=-1$ and $0 \leq \theta \leq 180^{\circ}$ is $\theta=135^{\circ}$.

If $\cos \theta-\sin \theta=\frac{1}{\sqrt{2}}$, then squaring both sides gives $\cos ^{2} \theta+\sin ^{2} \theta-2 \sin \theta \cos \theta=\frac{1}{2}$.

Using the identities $\cos ^{2} \theta+\sin ^{2} \theta=1$ and $2 \sin \theta \cos \theta=\sin 2 \theta$, this can be simplified to $\sin 2 \theta=\frac{1}{2}$.

We need $0 \leq \theta \leq 180^{\circ}$, which is equivalent to $0 \leq 2 \theta \leq 360^{\circ}$.

Since $\sin 2 \theta=\frac{1}{2}$, this means $2 \theta=30^{\circ}$ or $2 \theta=150^{\circ}$.

Thus, we get two additional solutions, $\theta=15^{\circ}$ and $\theta=75^{\circ}$.

Notice that $45^{\circ}<75^{\circ}<90^{\circ}$, so $\cos 75^{\circ}<\sin 75^{\circ}$. This means $\cos 75^{\circ}-\sin 75^{\circ}<0$ and cannot equal $\frac{1}{\sqrt{2}}$.

Thus, we reject the answer $75^{\circ}$. It can be shown that the other two solutions are indeed solutions.

ANSWER: $15^{\circ}, 135^{\circ}$