Solve: $-\frac{1}{\sqrt{2}} < \sin \theta + \cos \theta < \frac{1}{\sqrt{2}}$ The question is:

Solve $$-\frac{1}{\sqrt{2}} \lt \sin\theta + \cos\theta < \frac{1}{\sqrt{2}}$$ for values of $\theta$ between $0^\circ$ and $180^\circ$.

I realized that:
$$\begin{align}
-\frac{1}{\sqrt{2}} < \sin\theta + \cosθ &< \frac{1}{\sqrt{2}} \\[4pt]
\left|\sin\theta + \cosθ\right| &< \frac{1}{\sqrt{2}} \\[4pt]
\left(\sin\theta + \cos\theta\right)^2 &< \frac12 \\[4pt]
\sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta &< \frac12 \\[4pt]
1 + 2\sin\theta\cos\theta &< \frac12 \\[4pt]
1 + \sin 2\theta &< \frac12 \\[4pt]
\sin 2\theta &< -\frac12
\end{align}$$ 
But I don't know where to go from here. Can someone help me figure out how to get to the answer in the book: $105^\circ < θ < 165^\circ$. Thanks!

From that point you can do
$$\sin2\theta<-\frac12$$
Since $\theta$ is between $0$ and $180^o$, $2\theta$ is between $0$ and $360^o$. Then,
$$210^o<2\theta<330^o$$
Now, dividing by $2$, you obtain your book's result.
