<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] We have $$\sin(\theta) + \cos(\theta) = 1 \text{ and } \sin^2(\theta) + \cos^2(\theta) = 1$$
If we have $$a+b = 1 \text{ and } a^2 + b^2 =1 \tag{$\star$}$$then either $a=0,b=1$ or $a=1, b=0$. This gives us $$\sin(\theta) = 0, \cos(\theta) = 1 \text{ (or) }\sin(\theta) = 1, \cos(\theta) = 0$$Hence, $$\theta = 0^{\circ},90^{\circ}$$

EDIT(Expanding out the implications of $(\star)$).
We shall show here that $(\star) \implies$ either $a=0,b=1$ or $a=1, b=0$.
We have $a=1-b$. Plugging this in $a^2 + b^2 = 1$, we get that $$(1-b)^2 + b^2 = 1 \implies 1-2b+b^2 + b^2 = 1 \implies 2b^2-2b = 0 \implies 2b(b-1) = 0$$
Hence, $b=0$ or $b=1$. Plugging in $b=0$ in $a=1-b$ gives us $a=1$. Similarly, plugging in $b=1$ in $a=1-b$ gives us $a=0$. Hence, the possible solutions are
$$(a,b) = (0,1) \text{ or }(1,0)$$
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