# Trigonometric Equation

1. Jan 15, 2005

Hello all

Just refreshing in maths, and want to know if I am doing this correctly:

Solve the equation $$\sin x = 0.2$$ on the interval $$0 \leq x < 2\pi$$.

I took $$\arcsin(0.2)$$, however how do you solve for all solutions in the interval? (i.e what if the interval had been $$0 \leq x < \pi$$)

Thanks

2. Jan 15, 2005

### dextercioby

You know that "arcsin" only gives you the solution on the interval $[-\frac{\pi}{2},\frac{\pi}{2}]$,out of which you will only find useful the positive part.Okay,so that leaves you with another 3 intervals of width $\frac{\pi}{2}$.

HINT:"arccos" will be useful for $[\frac{\pi}{2},\pi]$.
Use that
$$\cos(\frac{\pi}{2}-x) =\sin x$$

Daniel.

PS.What about the interval $(\pi,2\pi)$ ??

3. Jan 15, 2005

so $$\cos(\frac{\pi}{2}-x) =\sin x$$

then $$\arccos(\frac{\pi}{2}-x) = 0.2$$ and then just solve for x.

for $[{\pi},[\frac{3\pi}{2}]$

and $[\frac{3\pi}{2}],[\frac{2\pi}]$ would i employ the same reasoning?

Thanks

4. Jan 15, 2005

### dextercioby

$$\cos(\frac{\pi}{2}-x)=\sin x=0.2$$

Solve for "x".

Daniel.

PS.On the second interval,the equation does not have solutions.Can u see why??

5. Jan 15, 2005

hmmm...isn't there an easier way to do this? Shouldn't I just be able to use the period of $$\sin x$$ which is $$2\pi$$ and divide it into 4 equal parts?

Thanks

6. Jan 15, 2005

### dextercioby

Believe,me there ain't no simpler method.2 simple equations solvable via "arcsin" and "arccos" and the fact that,on the interval $(\pi,2\pi)$ the "sine" is negative will give you the answer.

Daniel.

7. Jan 15, 2005

### Curious3141

Just use the fact that $$\sin x$$ is positive for $$x$$ in the 1st and 2nd quadrants. That means a domain of $$0 \leq x < \pi$$

By plugging in $$\arcsin(0.2)$$, you'll be finding the 1st quadrant value. To get the other value in the second quadrant, just subtract the value of $$x$$ from $$\pi$$, et voila ! That is, if $$x$$ is one answer, then $$(\pi - x)$$ is the other required answer.

Another way of looking at this is :

$$\sin (x) = \sin (\pi - x)$$

Equivalent expressions for cosine and tangent are :

$$\cos(x) = \cos (2\pi - x) = \cos (-x)$$

$$\tan(x) = \tan (\pi + x)$$

Last edited: Jan 15, 2005
8. Jan 15, 2005

### dextercioby

Did u have a rough night???
1.There's no such thing as $\arcsin 2$

2.If he adds \pi to the first solution,he'll end up in the third quadrant...

3.$$\sin(\pi+x)=-\sin x$$.

If u don't know,at least don't show it...

Daniel.

EDIT:And one more thing:it's "VOILÀ",or "voilà"...

Last edited: Jan 15, 2005
9. Jan 15, 2005

### Curious3141

I made a typo, genius. I just got up, Singapore time and everything.

I know how to do this, you obviously don't. Honestly, WTF are you doing asking him to consider the arccosine in such a trivially simple problem ?!

If YOU don't know, don't harangue others who do.

10. Jan 15, 2005

### dextercioby

You must have had a rough night... :tongue2: Was it because in Singapore the air it's more humid,hence it is harder to breathe??

You know them alright,you edited your post twice until got the final result.The last time was after i posted the reply in which i showed the mistakes...

Daniel...

PS.Expressions as "WTF" denote your level of intelligence... :yuck:

11. Jan 15, 2005

### Curious3141

OK, that's it !

FYI, I had just finished editing my post when you had waltzed in with you arrogant little "corrections". I don't need *you* to teach *me* trig.

It's obvious that my knowledge of trig and possibly other areas of math exceeds yours. Might I remind you of that thread you started in the brainteaser section with a (to me) simple question, which I summarily solved, yet which you couldn't crack with your wonderful noggin ?

I can see you are an arrogant, small minded phallus. You didn't bother to acknowledge my solution of your poser, because you're bearing a grudge after our run in about fundamental "axioms" in Physics. Which, BTW, I have yet to hear your learned dogma on.

I can assure you my IQ is at least as high, if not higher than yours. Currently, I'm not even doing Maths/Physics/Engg as a primary discipline. My knowledge was gained many years back in school, with no practice in between. I think it's pretty good considering that.

Keep your petty little taunts to the grade school yard where you belong.

12. Jan 15, 2005

### Integral

Staff Emeritus
opps wrong button, The little red triange at the bottom left is the report message button.

Both of your are getting a bit carried away. Just stop, I think the question has been answered.

Last edited: Jan 15, 2005
13. Jan 15, 2005

if we get the value of $$x$$ in the first two quadrants, that means there are no solutions in the third and fourth quadrants?

Thanks

14. Jan 15, 2005

### Curious3141

Correct, there are no solutions for this equation in the 3rd and 4th quadrants.

15. Jan 15, 2005

### dextercioby

Yes,simply because the "sine" is negative,and your equation is asking for solutions which make "sine" 0.2,which is positive.

Daniel.

16. Jan 15, 2005