Trigonometry equation How would I figure out the following trig equation.

$$\cos (x)(\csc x-\sqrt{2})=0$$

I know that $\csc(x)=\sqrt{2}$ is $\,\large\frac{\pi}{4}\,$ and $\,\large\frac{3 \pi}{4}\,$
So would I take $\cos$ of these two?
 A: In the given equation, $\cos (x)$ is a factor of $\,cxc(x) - \sqrt 2)$; the expression does not ask you to determine $\,cxc(x) - \sqrt 2)$ as the argument of the $\cos$ function. 
So we check when each factor equals $0$:
$$\cos(x)(\csc(x)-\sqrt{2})=0 \implies \cos x = 0\;\;\text{or}\;\;\csc x -\sqrt 2 = 0 \implies \csc x = \sqrt 2$$
Yes indeed, the solution to $\csc(x)=\sqrt{2}$ is $x = \dfrac{\pi}{4}$ and $x = \dfrac{3 \pi}{4}$
So you've found two of four solutions for $x \in [0, 2\pi)$.
Now you need to only to determine when $\bf \cos x = 0$.
A: So, either $\cos x=0\iff x=(2m+1)\frac\pi2$ where $m$ is any integer.
$0\le (2m+1)\frac\pi2< 2\pi\implies 0\le 2m+1<4\implies m=0,1$
or, $\csc x=\sqrt2\implies \sin x=\frac1{\sqrt2}=\sin \frac\pi4$
So in that case, $x=n\pi+(-1)^n \frac\pi4$ where $n$ is any integer.
If $n$ is even $=2r$(say) $x=2r\pi+\frac\pi4\implies 0\le 2r\pi+\frac\pi4<2\pi\implies r=0 $
If $n$ is odd $=2r+1$(say) $x=(2r+1)\pi-\frac\pi4\implies 0\le (2r+1)\pi-\frac\pi4<2\pi\implies r=0 $
A: Use the fact that $AB = 0 \implies A = 0 \text{ or } B = 0$.
You have already found when $\csc(x) = 0$, so $\frac{\pi}{4}$ and $\frac{3\pi}{4}$ are both solutions.  Now, when does $\cos(x) = 0$?  When else is $\csc(x) = \sqrt{2}$?
