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

Another approach not so fast in this case but not yet mentioned is a general method to solve any linear equation in $\sin
\theta $ and $\cos \theta $ of the form  $a\sin \theta +b\cos \theta =c$. We express these functions in terms of $t=\tan \frac{\theta }{2}$:
$$
\begin{equation*}
\sin \theta =\frac{2t}{1+t^{2}}
,\qquad \cos \theta =\frac{1-t^{2}}{1+t^{2}}
\end{equation*}$$
and solve a quadratic equation in $t$. The equation $\sin \theta +\cos \theta =1$ is transformed in
$$
\begin{eqnarray*}
2t+1-t^{2}=1+t^{2}
&\Leftrightarrow &\left( t-1\right)t=0\Leftrightarrow t=1,t=0.
\end{eqnarray*}
$$
For $t=\tan \frac{\theta }{2}=1$ we get the solution $\theta =90{{}^\circ}$ and for $t=\tan \frac{\theta }{2}=0$ the solution  $\theta=0{{}^\circ}$.
