<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] $$1=\sin\theta+\cos\theta\implies\frac{1}{\sqrt 2}=\sin\theta\cos\frac{\pi}{4}+\sin\frac{\pi}{4}\cos\theta=\sin\left(\theta+\frac{\pi}{4}\right)\Longrightarrow$$
$$\theta+\frac{\pi}{4}=\begin{cases}\frac{\pi}{4}\\{}\\\pi-\frac{\pi}{4}=\frac{3\pi}{4}\end{cases}\implies\theta=0\;\;\vee\;\;\frac{\pi}{2}$$
If you don't like radians, change
$$\frac{\pi}{4}\sim45^\circ\;,\;\;\frac{3\pi}{4}\sim 135^\circ\;,\;\;\pi\sim 180^\circ$$
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