# Trigonometry Equation

1. Aug 6, 2008

### truewt

Okay guys, this equation is bothering me and I have no idea how to start:

$$4cosec(X) - 3sec(X) = 4cos(2X)$$

Last edited by a moderator: Aug 6, 2008
2. Aug 6, 2008

### jeffreydk

Is that first term $4\csc(x)$? And what are you trying to do with the equation, solve it?

3. Aug 6, 2008

### truewt

Yup, to both queries.

I tried many approaches but can't solve :(

4. Aug 6, 2008

### rock.freak667

write cosec(x) and sec(x) in terms of cos and sine, then bring the left side to a common denominator.

EDIT: wait...that might get complicated.

I doubt it has "nice" solutions though.

Last edited: Aug 6, 2008
5. Aug 6, 2008

### truewt

i tried that. It's a damn freaking weird question. I can't solve.

6. Aug 6, 2008

### truewt

what I will end up with is $$4cos(X) - 3sin(X) = sin(4X)$$

7. Aug 6, 2008

### snipez90

Unfortunately that's probably the nicest form you can get. The right hand side folds up nicely but then you'd have to deal with the 4cos(x) - 3sin(x). I tried expanding cos(2x) but that got nasty quick.

8. Aug 11, 2008

### The_ArtofScience

I don't think that statement is correct. There aren't any real solutions.

9. Aug 11, 2008

### tiny-tim

Hi truewt!

Hint: write it $$\frac{4}{5}cos(X) - \frac{3}{5}sin(X) = \frac{1}{5}sin(4X)$$

Then let sin(Y) = 4/5.

Does that help?

10. Aug 11, 2008

### DecayProduct

Being fairly new to trig, I'd like to ask a question based on this question. I'm a hands-on kind of guy, so I take something and just jam a number where x is to see what happens. Let's say we stick 30 (degrees) into it. The left side becomes:

4csc30 - 3sec30 or 4(2) - 3(1.15) = 4.55

The right side becomes:

4cos2(30) = 4(1-2sin^2(30)) = 2

Put them together and you get 4.55 = 2, which is never true. Additionally, I've gone through the trouble of defining everything in terms of sin x, and the two sides of the equation cannot equal. What are we looking for here?

11. Aug 11, 2008

### snipez90

DecayProduct, I assume we're trying to find all values of X that satisfies the equation.

Now if you're trying to imply that there are no real solutions based on substituting one value of X then that's not very reliable. Sure 4.55 =/= 2 but that doesn't really tell us anything except that if x = 30 deg then the equation is not satisfied.

You can only deduce that an equation is not satisfied if you can eliminate the variables and find a contradiction. But what you've done is basically the equivalent of taking any equation, say one involving polynomials, and plugging in a value for x, finding out the two sides don't equal, and claiming that the two sides are not equal. If you claim no solutions, you'll have to find a clear cut contradiction.

12. Aug 12, 2008

### xavier_r

I guess you can change the equation to,

$$(1/2)*cos^{-1}((4cosec(X) - 3sec(X))/4) = X$$
So let

$$f(X) = (1/2)*cos^{-1}((4cosec(X) - 3sec(X))/4)$$

Then use a computer program or any scientific calculator, to find the limit of $$f^n(X)$$ as n tends to infinity... where $$f^n(X) = f(f(f(... n times(f(f(X)))))$$

13. Aug 12, 2008

### NoMoreExams

Why did you pick 30 degrees? This isn't an identity that is asked to be shown but rather a "solve for x" type situation I believe.

14. Aug 12, 2008

### DecayProduct

30 wasn't all I tried. But it was arbitrary, to see what happens. Please forgive my ignorance when it comes to such things. I don't claim to know enough to say if the equation has a solution or not. But, even if you boil the whole thing down into the rudimentary parts, that is, x's, y's and r's, I can't get the two sides to balance. If you convert all the units in terms of sinx, I can't get it to balance. cos x = sin x(sqrt(1/sin^2 x -1)), and cos 2x = 1-2sin^2 x.

It is a messy bunch of scribbling I have here, but the left still doesn't equal the right. I can't figure how any angle would satisfy it. What gives?

15. Aug 12, 2008

### Avodyne

If you have a graphing calculator or computer program, you can just plot the original left side minus the original right side. You will then see that (in the range between 0 and 360 degrees) there are two real solutions.