<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] Using the fact that $\sin \theta+\cos \theta=\sqrt{2}\cdot\sin (\theta+\frac{\pi}{4})$, this problem is reduced to find an angle $\theta\in [0, 2\pi]$ such that 
$$\sin (\theta+\frac{\pi}{4})=\frac{\sqrt{2}}{2}.$$
As we know, if $\sin\alpha=\sqrt{2}/2$, it must be the case that $\alpha=\frac{\pi}{4}+2k_1\pi$ or $\alpha=\frac{3\pi}{4}+2k_2\pi$, where $k_1$ and $k_2$ can be any integer. In this problem, since $\theta\in [0,2\pi]$, we have 
$$\theta+\frac{\pi}{4}\in\bigg[\frac{\pi}{4}, \frac{9\pi}{4}\bigg].$$
Thus,the possible values of $k$'s are: $k_1=0$, $k_1=1$, $k_2=0$, which correspond to three angles in $[0, 2\pi]$: $\theta_1=\frac{\pi}{4}-\frac{\pi}{4}=0$, $\theta_2=\frac{9\pi}{4}-\frac{\pi}{4}=2\pi$, and $\theta_3=\frac{3\pi}{4}-\frac{\pi}{4}=\frac{\pi}{2}$.
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