Solving $\sin \frac{\theta}{2} + \cos \frac{\theta}{2} = \sqrt{2}$ Does anyone have some tips for me how to go about the problem in the image? 
$$\sin \frac{\theta}{2} + \cos \frac{\theta}{2} = \sqrt{2}$$
I know it's supposed to be simple, but I can't figure out why the solution is: 
$90^{\circ}+ 720^{\circ}k$ where $k=1,2,3,...$  
Intuitively, I understand the answer, but that's not the point, I would like to understand the correct mathematical solution. Thank you very much!
 A: 
Notice:
$$\cos\left(\frac{\theta}{2}\right)+\sin\left(\frac{\theta}{2}\right)=\frac{\sqrt{2}\cos\left(\frac{\theta}{2}\right)}{\sqrt{2}}+\frac{\sqrt{2}\sin\left(\frac{\theta}{2}\right)}{\sqrt{2}}=\sqrt{2}\left[\frac{\cos\left(\frac{\theta}{2}\right)}{\sqrt{2}}+\frac{\sin\left(\frac{\theta}{2}\right)}{\sqrt{2}}\right]=$$
$$\sqrt{2}\left[\sin\left(\frac{\pi}{4}\right)\cos\left(\frac{\theta}{2}\right)+\cos\left(\frac{\pi}{4}\right)\sin\left(\frac{\theta}{2}\right)\right]=\sqrt{2}\sin\left(\frac{\pi}{4}+\frac{\theta}{2}\right)$$

So:
$$\cos\left(\frac{\theta}{2}\right)+\sin\left(\frac{\theta}{2}\right)=\sqrt{2}\Longleftrightarrow$$
$$\sqrt{2}\sin\left(\frac{\pi}{4}+\frac{\theta}{2}\right)=\sqrt{2}\Longleftrightarrow$$
$$\sin\left(\frac{\pi}{4}+\frac{\theta}{2}\right)=1\Longleftrightarrow$$
$$\frac{\pi}{4}+\frac{\theta}{2}=\frac{\pi}{2}+2\pi n\Longleftrightarrow$$
$$\frac{\theta}{2}=\frac{\pi}{4}+2\pi n\Longleftrightarrow$$
$$\theta=\frac{\pi}{2}+4\pi n$$
Where $n\in\mathbb{Z}$.
A: Set $X=\cos(\theta/2)$ and $Y=\sin(\theta/2)$; then you have
$$
\begin{cases}
X+Y=\sqrt{2} \\[4px]
X^2+Y^2=1
\end{cases}
$$
Instead of solving for $Y$ in the first equation and substituting, which works, you can notice that $X^2+Y^2=(X+Y)^2-2XY$, so the second equation becomes
$$
2-2XY=1
$$
and you finally have
$$
\begin{cases}
X+Y=\sqrt{2} \\[4px]
XY=1/2
\end{cases}
$$
so the solutions can be found with the roots of
$$
z^2-\sqrt{2}{z}+\frac{1}{2}=0
$$
This is $(z-\frac{1}{\sqrt{2}})^2=0$, so we conclude
$$
X=Y=\frac{1}{\sqrt{2}}
$$
and finally
$$
\frac{\theta}{2}=\frac{\pi}{4}+2k\pi
$$
A: Note that: $$\sin \frac{\theta}{2} + \cos \frac{\theta}{2} = \sqrt{2} \cos \left(\frac{\theta}{2} - \frac{\pi}{4} \right) \quad \quad \quad \quad (\star)$$
So your equation reduces down to $\cos \left(\frac{\theta}{2} - \frac{\pi}{4}\right) = 1 \iff \frac{\theta}{2} - \frac{\pi}{4} = 2\pi k$ where $k$ ranges over the integers. 
This can be re-written equivalently as $\theta = \frac{\pi}{2} + 4\pi k$ or in degrees: $\theta = 90^{\circ} + 720^{\circ} k$. 

$(\star):$ Assume that $\sin x + \cos x = R\cos (x - \alpha)$ then expanding the left hand, we have $\sin x + \cos x = R\cos x \cos \alpha + R\sin x \sin \alpha$. Now we have $R\cos \alpha = 1$ and $R\sin \alpha = 1$, squaring and adding gives us $R^2 (\cos^2 \alpha + \sin^2 \alpha) = 2 \Rightarrow R = \sqrt{2}$. Dividing the two equations gives $\frac{R\sin \alpha}{R\cos \alpha} = \tan \alpha = 1\Rightarrow \alpha = \frac{\pi}{4}$, whence $(\star)$. 
A: Firstly, denote $x=\frac{\theta}{2}$. Let $t=\tan \frac{x}{2}$. The equation $\sin x + \cos x = \sqrt 2$ became $$\frac{2t}{1+t^2} + \frac{1-t^2}{1+t^2}=\sqrt 2.$$
Therefore $2t + 1 -t^2 = \sqrt 2(1+t^2) \Rightarrow (\sqrt 2+1)t^2 -2t + \sqrt 2 - 1 = 0$. This equation has the solutions $t_1=t_2=\sqrt 2 - 1$. So
$$\frac{x}{2}\in \{\arctan(\sqrt 2-1)+k\pi\;|\;k\in\mathbb Z\}.$$
Thus, since $\theta = 4x$, we get
$$\theta \in \{4\arctan(\sqrt 2 - 1)+4k\pi\;|\;k\in\mathbb Z\}.$$
We have $$\tan(2\arctan(\sqrt 2-1))=\frac{2(\sqrt 2-1)}{(\sqrt 2-1)^2-1}=1,$$
hence $2\arctan(\sqrt 2-1)=\frac{\pi}{4} \Rightarrow 4\arctan(\sqrt 2-1)=\frac{\pi}{2}$. In conclusion,
$$\theta \in \{\frac{\pi}{2}+4k\pi\;|\;k\in\mathbb Z\}.$$
A: $$ \sin \frac{\theta}{2} + \cos \frac{\theta}{2} = \sqrt{2} $$
Squaring,
$$ 1+ \sin \theta = 2 $$
$$\sin \theta = 1 $$
$$ \theta = \pi/2  + 2k \pi. $$
EDIT1: 
The squaring has introduced spurious roots as expected. We need to take one more turn of radius vector. 
$$ \theta = \pi/2  + 4\, k \pi. $$
EDIT2:
First answer was okay. Since only one solution is found uniquely upto cycle period, nothing spurious got in, so please ignore EDIT1. Even otherwise,
$$  \frac{\sin \theta/2}{\sqrt{2} } +\frac{\cos \theta/2}{\sqrt{2} } = 1= \sin \pi/2 $$
$$  \sin (\theta/2 + \pi/4)= \sin \pi/2 $$
$$(\theta/2 + \pi/4)= \pi/2  + (-1)^k  \cdot k \pi $$
$$ \theta = \pi/2  + 2k \pi, $$
where k is an integer, positive or negative.
