Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

3 Answers 3

up vote 5 down vote accepted

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 $\bf \cos x = 0$.

share|improve this answer
    
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 $

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.