<s>[INST] Solving $\sin \theta + \cos \theta=1$ in the interval $0^\circ\leq \theta\leq 360^\circ$ 
Solve in the interval $0^\circ\leq \theta\leq 360^\circ$ the equation $\sin \theta + \cos \theta=1$.

I've got the two angles in the interval to be $0^\circ$ and $90^\circ$, it's not an answer I'm after, I'd just like to see different approaches one could take with a problem like this.
Thank you!
Sorry, my approach: 
$$\begin{align}
\frac{1}{\sqrt 2}\sin \theta + \frac{1}{\sqrt 2}\cos \theta &= \frac{1}{{\sqrt 2 }} \\
\cos 45^\circ\sin \theta + \sin 45^\circ\cos \theta &= \frac{1}{\sqrt 2} \\
\sin(\theta + 45^\circ) &= \frac{1}{\sqrt 2} \\
\theta + 45^\circ &= 45^\circ,\ 135^\circ \\
\theta &= 0^\circ, \ 90^\circ
\end{align}$$
 [/INST] I'd write $\sin \theta + \cos \theta = \sqrt{2} \sin \left(\theta + \dfrac{\pi}{4}\right)$ and go from there.
A similar tactic works for all equations of the form $a \sin \theta + b \cos \theta = c$ for constant $a,b,c$.
</s>