Radians/Trigonometric Equations

[JBD2]

Homework Statement

Solve for x, where 0\leq x < 2\pi. Then give a general solution.
The question:
sin\²x-1=0

Homework Equations

Perhaps sin\theta=\frac{y}{r}?
None else that I'm aware of.

The Attempt at a Solution

So far I obviously have sin\²x=1. Using a graphing calculator I have solved the problem, however I am unsure how to solve this algebraically. With that, I leave you with my question.

 

[happyg1]

where is \frac{y}{r}=1?
I'm looking at the unit circle...

 

[JBD2]

Well I just mean \frac{y}{r} is \frac{opposite}{hypotenuse}. What is the unit circle?

In general I'm just finding it difficult to do this type of question as the textbook is not explaining it very clearly. The answer in the back is:

\frac{\pi}{2},\frac{3\pi}{2};(2n+1)\frac{\pi}{2}, n any integer

 

[Dick]

Look at a graph of sin. It's periodic with period 2*pi meaning that the value of sin(x) and sin(x+2*pi) are equal. The only solution in [0,2*pi] is pi/2. The other solutions are just periodic extensions of that. And 3*pi/2 isn't one of them. What's up with that?

 

[JBD2]

What do you mean by the solution for [0,2\pi]? I kind of understand where you explained that \frac{\pi}{2} was the solution of where sinx=1, but I'm not sure like you said how is \frac{3\pi}{2} an answer? \frac{\pi}{2} is equivalent to 90° which satisfies sin\theta=1, but \frac{3\pi}{2} is equivalent to 270° which would be -90°, which wouldn't be a solution right?

PS I'm sure I might be repeating a lot of what you said, I'm just trying to make sure that I understand correctly.

 

[Dick]

Well I just mean \frac{y}{r} is \frac{opposite}{hypotenuse}. What is the unit circle?

In general I'm just finding it difficult to do this type of question as the textbook is not explaining it very clearly. The answer in the back is:

\frac{\pi}{2},\frac{3\pi}{2};(2n+1)\frac{\pi}{2}, n any integer

I was just pointing out that the answer in the back is not correct. E.g. 3*pi/2 is WRONG.

 

[JBD2]

So would that mean that (2n+1)\frac{\pi}{2} is wrong? Because if:

Let n be 1
(2(1)+1)\frac{\pi}{2}
(2+1)\frac{\pi}{2}
\frac{3\pi}{2}

Then it appears to be correct

 

[arildno]

Are you sure the equation given in the book wasn't:
\sin^{2}(x)-1=0
In that case, the answers in your book is correct.

 

[JBD2]

Oh wow you're exactly right, I'm sorry about that, I guess I can't re-edit my original post.

How would I go about solving this?

 

[arildno]

Well, that equation implies that:
\sin(x)=\pm{1}
Thus, you must combine the solutions of the two equations:
\sin(x)=1 and \sin(x)=-1 in order to get the full solution set.

 

[JBD2]

Oh ok so that's why there would be 90° and -90°?

 

[arildno]

That's correct!

 

[JBD2]

Ok thanks, and I'm really sorry that I screwed up on that, I'll be sure to check harder next time. Thanks.

 

[JBD2]

Oh one more question, I don't understand how (2n+1) becomes part of the answer, can someone explain? Thank you.

 

[Dick]

The x values where sin(x)^2=1 are pi/2, 3*pi/2, 5*pi/2, 7*pi/2 etc etc. (2n+1) is a concise way of expressing the numerator of those values.

 

[JBD2]

Ok I see, and I figured I'd ask this instead of making a new thread. When trying to graph the functions on a graphing calculator (I use the ti-83+), how do I know what to make the window size (X scale/Y scale)? Because in my textbook it mentions things like "Since solutions are to be in the domain 0\leq x<2\pi, choose radian mode and use a window with Xmin = 0, Xmax = 2\pi, and Xscl (X Scale)=\frac{\pi}{2}. I just don't understand how they know what to set the x scale to so accurately. Thanks.

 

[JBD2]

Ok nevermind I figured it out on my own, sorry for this double post, I'm done with this thread.