trig problem

[Jake Bunce]

Directions:
Find all solutions in the interval (0, 2pi]

Problem:
tan^2(x)*csc(x) = tan^2(x)

Translated into English since it might look confusing typed out... the tangeant squared of x times the cosecant of x equals the tangeant squared of x. That is not the tangeant of (x^2).

I determined there were no solutions, but that wasn't the right answer. I still think I got it right. If I can get the points for this one problem, I will have an A instead of a B in that class for this past Spring semester.

Can you guys tell me what answer you get and how you worked it out?

[CeleronXT]

Directions:
Find all solutions in the interval (0, 2pi]

Problem:
tan^2(x)*csc(x) = tan^2(x)

Translated into English since it might look confusing typed out... the tangeant squared of x times the cosecant of x equals the tangeant squared of x. That is not the tangeant of (x^2).

I determined there were no solutions, but that wasn't the right answer. I still think I got it right. If I can get the points for this one problem, I will have an A instead of a B in that class for this past Spring semester.

Can you guys tell me what answer you get and how you worked it out?

I have no idea, but you could do what I do in these problems that take a lot of steps: write PHP calculators that does it all out for you.

[Jake Bunce]

It's not a very long problem... or, it isn't the way that I did it, which I think is right!

[Chen]

If you move the tg^2(x) to the other side you get this:
tg^2(x) * (cosec(x) - 1) = 0
So first answer, is tg(x) = 0, which happens when x is 0 or pi. So x1 = 0, x2 = pi.
The other answer is cosec(x) = 1, in other words sin(x) = 1, which happens when x is pi/2.

So the answers are 0, pi/2 and pi. But... because you have tg(x) it means cos(x) has to be different than 0, so x has to be different than pi/2 and 3pi/2. So we eliminate the pi/2 solution and you are left with 0 and pi.

[Jake Bunce]

When x is 0 or pi, the csc is undefined. But I guess 0 * undefined = 0 , so you're right. When I was doing this problem I immediately eliminated the solutions where csc is undefined... I didn't think about the tan being 0. I approached the problem from the standpoint that the only solutions had to make the csc equal 1, which makes the tan undefined.

NOOOOOOOO 89.7% in the class!

[Chen]

Well it depends on what class this is for... and what you've already covered. But if "no solutions" is not right, I think 0 and pi will do because I don't see any other possible answers other than this.

[Chen]

Technically, you can replace the cosec(x) with 1/sin(x) so you get this:
sin(x) / cos^2(x) = sin^2(x) / cos^2(x)
So once you eliminate the cos(x) = 0 solutions, you can make it:
sin(x) = sin^2(x)
Or:
sin(x) * [1 - sin(x)] = 0
So sin(x) equals 0 or 1, but it can't be 1 because cos(x) has to be different than 0.

[Beorn]

Directions:
Find all solutions in the interval (0, 2pi]

Problem:
tan^2(x)*csc(x) = tan^2(x)

Translated into English since it might look confusing typed out... the tangeant squared of x times the cosecant of x equals the tangeant squared of x. That is not the tangeant of (x^2).

I determined there were no solutions, but that wasn't the right answer. I still think I got it right. If I can get the points for this one problem, I will have an A instead of a B in that class for this past Spring semester.

Can you guys tell me what answer you get and how you worked it out?

Divide each side by tan^2(x):

csc(x) = 1

1/sin(x) = 1


Multiply each side by sin(x)

1 = sin(x)

Asin(1) = x

x = pi()

Agh...I've tried this two different ways, and both result in a null set...I'll ask my teacher tomorrow, we just had a test trigonometic derivations and identities a few days ago--literally.

Mike

[Jake Bunce]

Divide each side by tan^2(x):

csc(x) = 1

1/sin(x) = 1


Multiply each side by sin(x)

1 = sin(x)

Asin(1) = x

x = pi()

Agh...I've tried this two different ways, and both result in a null set...I'll ask my teacher tomorrow, we just had a test trigonometic derivations and identities a few days ago--literally.

Mike

When you multiply or divide both sides by a variable quantity you can introduce extraneous solutions, so be careful.

[Jake Bunce]

Wait, that's wrong anyways.

x = arcsin(1)
x = 90 deg or pi/2

Besides, arcsin can only return values between -pi/2 and pi/2 . It can't return pi. That's another thing, you have to be careful when coverting to arc functions.

[Chen]

Beorn, you can't just divide by tg^2(x) because you lose possible solutions, and when you multiply you add new foreign solutions.

[Beorn]

Beorn, you can't just divide by tg^2(x) because you lose possible solutions, and when you multiply you add new foreign solutions.

True....that was the second way I tried it....first was to factor out sin(x)...and that wasn't pretty....But, I'm surprised no one has asked this question: Is Jakeman sure he got it wrong?

[Jake Bunce]

Is Jakeman sure he got it wrong?

My teacher is sure I got it wrong.

I'm still questioning his decision though. The only possible answers are 0 deg, 90 deg, and 180 deg. 90 deg results in undefined terms on both sides, so that's out. 0 and 180 deg result in an undefined term on the left side, but it's cancelled out by a zero term, so both sides equal zero.

I emailed him asking him if the 0 and 180 deg solutions were correct, and if an undefined term in an equation is legitimate when it's cancelled by a zero term. I haven't received a reply yet.

[etones]

Divide each side by tan^2(x):

csc(x) = 1

1/sin(x) = 1


Multiply each side by sin(x)

1 = sin(x)

Asin(1) = x

x = pi()

Agh...I've tried this two different ways, and both result in a null set...I'll ask my teacher tomorrow, we just had a test trigonometic derivations and identities a few days ago--literally.

Mike

In trig you cant simply divide through as you loose possible answers (in real trig, there arent only 360 degerss [2 pi??]), you need to have a look at your trig substitutions, i think chen hit the nail on the head though

Then again, been a few years since my alevel maths days

Taz

[webdesigner]

brain overload, i'm only in grade 9