# 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?

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I do not understand the question. – Ron Gordon Feb 26 '13 at 17:28
sorry i have to find all solutions from o to 2pi – Fernando Martinez Feb 26 '13 at 17:29

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 $\bf \cos x = 0$.

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it is it when pi/2 and 3pi/2 – Fernando Martinez Feb 26 '13 at 17:31
Yes, indeed! So there are four solutions...to the given equation. The equation is true when either factor $\;\cos x\;\;$ or $\;\csc x - \sqrt 2\;\;$ is equal to $0$. – amWhy Feb 26 '13 at 17:33
yes I get I understand now. – Fernando Martinez Mar 2 '13 at 19:00

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) $x=(2r+1)\pi-\frac\pi4\implies 0\le (2r+1)\pi-\frac\pi4<2\pi\implies r=0$

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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 $\csc(x) = \sqrt{2}$?

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