How to solve $\sin x +\cos x = 1$? No matter how I do it, I always end up with $x = 0, 90, 270$ and $360$. All of those except $270$ is right, but I can't quite figure out how to get the $270$ degrees out of the answer. I've tried using trig identities, I've tried squaring both sides, but I always end up with $$2\sin x\cos x$$ which then leads me to $x = 0, 90, 270, 360$. But $$\sin (270) + \cos (270) = -1$$ so I'm doing something wrong.
 A: The steps you followed are perfectly correct!  The only thing you have to do that you haven't done already is check for extraneous solutions. Whenever we square both sides, there is the chance that we get more solutions than we're looking for.  
So, you have correctly deduced that the list $0,90,270,360$ (all angles in degrees) contains all potential solutions.  After checking, you've noticed that $270$ is not a solution, but the rest are. So, the solutions are $0,90,$ and $360$.
As for why we get extra solutions: notice that although $270$ does not satisfy the original equation, we have
$$
(\sin(270)+\cos(270))^2=(-1)^2=1
$$
Which makes sense, since we just solved the equation 
$$
(\sin(x) + \cos(x))^2 = 1
$$
A: Another way to look at this is to consider the equation as representing the intersection of two "curves" in polar coordinates, one being $ \ r = 1 \ $ (the unit circle), the other being the line  $ \ \sin \theta \ + \ \cos \theta \ = \ 1 \ \ \Rightarrow \ \ r \sin \theta \ + \ r \cos \theta \ = \ r \ \ \Rightarrow \ \ r \ = \ x + y \ , $  with $ \ r \ $ set equal to 1 .  The line has intercepts at  ( 1 , 0 )  and  ( 0 , 1 )  , meeting the circle at  $  \ \theta = 0º \ \ \text{and} \ \ \theta = 90º $ .  (360º is considered to be merely another "angle-name" for  0º , so it is not really a distinct solution.)
A: Use $a\sin x + b\cos x = \sqrt{a^2+b^2}\cos(x- \tan^{-1}(\frac ba))$
Your problem:

$\sqrt2\cos(x-\frac\pi4) = 1$
$\cos(x-\frac\pi4) = \frac1{\sqrt2}$
$x - \frac\pi4 = 2n\pi \pm \frac\pi4$

Solving this gives 0,90,360 as solution.
A: Hint. Square both sides to get $$(\sin x+\cos x)^2=1+2\sin x\cos x=1+\sin 2x=1^2=1\implies\sin 2x=0.$$ The important thing here is to notice $2\sin x\cos x=\sin 2x$. Notice I wrote an implication, not an equivalence, so we can get some extraneous solutions at the end. A routine check can spot all of them.
A: Avoid squaring wherever practicable as it immediately introduces extraneous roots 
Using  Weierstrass substitution we have  $$\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}=1$$ where $t=\tan\frac x2$
$$\implies 2t+1-t^2=1+t^2\iff t^2-t=0\iff t=1,0$$
If  $\displaystyle\tan\frac x2=0\iff \frac x2=n180^\circ\iff x=n360^\circ$ where $n$ is any integer 
If  $\displaystyle\tan\frac x2=1\iff \frac x2=m180^\circ+45^\circ\iff x=m360^\circ+90^\circ$ where $m$ is any integer 
A: Using Double-Angle Formulas,
$$\sin x+\cos x=1\implies 2\sin\frac x2\cos\frac x2=1-\cos x=2\sin^2\frac x2$$
$$\implies \sin\frac x2\left(\cos\frac x2-\sin\frac x2\right)=0$$
$$(i)\sin\frac x2=0\implies \frac x2=n\pi\text{ where }n \text{ is any integer}$$
$$(ii) \cos\frac x2-\sin\frac x2=0\implies\cos\frac x2=\sin\frac x2\iff \tan\frac x2=1$$ Find the rest in my other answer
A: Hint: 
Use the formula $\sin(A+B) \equiv \sin A \cos B + \sin B \cos A$ to write $\sin x + \cos x$ in the form $R\sin(x+\alpha)$, where $R$ and $\alpha$ are numbers that you need to find. 
Once you have your $R$ and $\alpha$, simply solve $R\sin(x+\alpha)=1$.
A: $\sin x + \cos x = 1 \rightarrow (\sin x + \cos x )^2=1^2 \iff \color{green}{\underbrace{(\sin^2 x + \cos^2 x)}_{=1}}+2\sin x \cos x =\color{green}{1} \iff \sin x \cos x = 0  $
Now look at the unit circle to see when this is true. After you have found the values of $x$ for which this holds, be sure to check the outcome of $\sin x + \cos x$ for each of these values, to eliminate the outcomes that generate value $-1$ (these values of $x$ came in because of the squaring - but are not truly solutions).
