Solve $\sin x - \cos x = -1$ for the interval $(0, 2\pi)$ We have an exam in $3$ hours and I need help how to solve such trigonometric equations for intervals.
How to solve
$$\sin x - \cos x = -1$$
for the interval $(0, 2\pi)$.

Method $\#1$ 
Avoid squaring which immediately introduces extraneous roots which demand exclusion 
We have $\displaystyle\sin x-\cos x=-1$
$$\iff\sin x=-(1-\cos x)\iff2\sin\frac x2\cos\frac x2=-2\sin^2\frac x2$$
$$\iff2\sin\frac x2\left(\cos\frac x2+\sin\frac x2\right)=0$$
If $\displaystyle \sin\frac x2=0,\frac x2=n\pi\iff x=2n\pi$ where $n$ is any integer
If $\displaystyle\cos\frac x2+\sin\frac x2=0\iff\sin\frac x2=-\cos\frac x2$
$\displaystyle\iff\tan\frac x2=-1=-\tan\frac\pi4=\tan\left(-\frac\pi4\right)$
$\displaystyle\iff\frac x2=m\pi-\frac\pi4\iff x=2m\pi-\frac\pi2$ where $m$ is any integer
Method $\#2$
Let $\displaystyle1=r\cos\phi,-1=r\sin\phi\  \ \ \  (1)$ where $r>0$
$\displaystyle\cos\phi=\frac1r>0$ and $\displaystyle\sin\phi=-\frac1r<0$
$\displaystyle\implies\phi$ lies in the fourth Quadrant
On division, $\displaystyle\frac{r\sin\phi}{r\cos\phi}=-1\iff\tan\phi=-1$
$\displaystyle\implies\phi=-\frac\pi4$
$\displaystyle\sin x-\cos x=-1\implies r\cos\phi\sin x+r\sin\phi\cos x=r\sin\phi$
$\displaystyle\implies\sin(x+\phi)=\sin\phi$
$\displaystyle\implies x+\phi=r\pi+(-1)^r\phi$ where $r$ is any integer
If $r$ is even $=2a$(say) $\displaystyle\implies x=2a\pi$
If $r$ is odd $=2a+1$(say) $\displaystyle\implies x+\phi=(2a+1)\pi-\phi\iff x=2a\pi+\pi-2\phi$
Method $\#3$
Using Weierstrass Substitution,
$\displaystyle\frac{2u}{1+u^2}-\frac{1-u^2}{1+u^2}=-1\ \ \ \ (2)$ where $u=\tan\frac x2$
$\displaystyle\iff2u^2+2u=0\iff u(u+1)=0$
If $\displaystyle u=0,\tan\frac x2=0\iff\frac x2=b\pi$ where $b$ is any integer
If $\displaystyle u=-1,\tan\frac x2=-1$ which has been addressed in Method $\#1$ 
