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 $(0, 2\pi)$.
 A: HINT: Square both sides and you obtain $\sin 2x=0$ and then...
A: Square it. You get
$$
1 - 2 \sin x \cos x = \sin^2x + \cos^2 x - 2 \sin x \cos x = (\sin x - \cos x)^2 = 1
$$
which implies $\sin x = 0$ or $\cos x = 0$. Therefore the only possible solutions are multiples of $\pi/2$. See which ones are actual solutions of your problem (we squared, so $\sin x - \cos x = \pm 1$ when $x = k \pi / 2$). (Only $3 \pi / 2$ works in the interval $]0,2\pi[$.)
Hope that helps,
A: $$\sin x-\cos x=-1\\
\implies (\sin x-\cos x)^2=\sin^2x+\cos^2x-2\sin x\cos x=1+\sin(2x)=1$$
So,
$$\sin(2x)=0$$
Therefore, 
$$x=3\pi/2,\mbox{ as }x\in(0,2\pi)$$
A: Rewrite the equation as $\sin x=\cos x-1$, square both sides, and use the identity $\sin^2x=1-\cos^2x$ to get
$$1-\cos^2x=\cos^2x-2\cos x+1$$
which simplifies to
$$\cos x(1-\cos x)=0$$
so that either $\cos x=0$ or $\cos x=1$.  The former corresponds to $x=\pi/2$ and $3\pi/2$ in the interval $0\lt x\lt2\pi$.  The latter corresponds to $x=0$ or $2\pi$, neither of which is in the interval.  So there are just two solutions to the squared equation in the interval $(0,2\pi)$.  However, $x=\pi/2$ does not satisfy the original equation.  So there is just one solution, namely $x=3\pi/2$.
Added later:  Here's a second approach to solving the equation.  First, rewrite it as
$$\sin\theta=\cos\theta-1$$
and then think of this as describing the intersection of the unit circle in the $xy$ plane with the line
$$y=x-1$$
If you sketch this, you see that the line passes through the unit circle at $(x,y)=(0,-1)$ and $(1,0)$.  The first point corresponds to the angle $\theta=3\pi/2$, which is in the interval $(0,2\pi)$, while the other point corresponds to $\theta=0$, which isn't.
A: Hint: Use the formula for the sine of a difference to get
$$
\begin{align}
\sqrt2\sin(x-\pi/4)
&=\sqrt2\Big(\sin(x)\cos(\pi/4)-\cos(x)\sin(\pi/4)\Big)\\
&=\sin(x)-\cos(x)
\end{align}
$$
A: Use the fact that $\sin x = \frac{e^{ix}-e^{-ix}}{2i}$ and $\cos x = \frac{e^{ix}+e^{-ix}}{2}$ to rewrite the equation as (after simplifying)
$$e^{2ix} -(1-i)e^{ix}-(1+i)=0$$
Viewing this as a quadratic equation in $e^{ix}$, use the quadratic formula to get
$$e^{ix}= \frac{(1-i)\pm(1+i)}{2}$$
$$e^{ix} = 1 \text{ or } e^{ix} = -i$$
Take logs to get all solutions
$$x= 2\pi k \text{ or } x=\frac{3\pi}{2}+ 2\pi k$$
for $k\in\mathbb Z$. Restricting to $(0,2\pi)$ as required in the problem statement, we get only the solution
$$\boxed{x=\dfrac{3\pi}{2}}.$$
A: 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 $\#1$ 
A: Since $\cos(\pi/4)=\sin(\pi/4)=\sqrt{2}/2$ we have using the angle sum theorem
$$\sin(x)\sin(\pi/4)-\cos(x)\cos(\pi/4)=-\sqrt{2}/2\iff\cos(x+\pi/4)=\sqrt{2}/2$$
$$\iff
x+\pi/4=\pi/4\lor x+\pi/4=7\pi/4\iff x=3\pi/2$$
as $x\in(0,2\pi)$.
