<s>[INST] 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!
 [/INST] Hint: $$\sin \theta + \cos \theta \equiv \sqrt{2} \sin \left(\theta + \frac{\pi}{4}\right).$$
Or alternatively, in degrees $$\sqrt{2} \sin \left(\theta + 45^{\circ}\right).$$

Using your method, however, we get that $$\sin 2\theta < -
\frac{1}{2} \implies 210^{\circ} < 2\theta < 330^{\circ}.$$
</s>