algebra precalculus - Solving $sin theta + cos theta=1$ in the interval $0^circleq thetaleq 360^circ$

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:

$$\begin{align} \frac{1}{\sqrt 2}\sin \theta + \frac{1}{\sqrt 2}\cos \theta &= \frac{1}{{\sqrt 2 }} \\ \cos 45^\circ\sin \theta + \sin 45^\circ\cos \theta &= \frac{1}{\sqrt 2} \\ \sin(\theta + 45^\circ) &= \frac{1}{\sqrt 2} \\ \theta + 45^\circ &= 45^\circ,\ 135^\circ \\ \theta &= 0^\circ, \ 90^\circ \end{align}$$

Answer

A slightly 'expanded-upon' version of user67418's answer:

The circle here represents the parametric curve $(x=\cos\theta, y=\sin\theta)$, and the line is the line $x+y=1$, so their points of intersection are the points where $\cos\theta+\sin\theta=1$; at least for me, this is the clearest way of seeing that there are only the two solutions already mentioned.