- #1

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

[tex] 4cosec(X) - 3sec(X) = 4cos(2X)[/tex]

[tex] 4cosec(X) - 3sec(X) = 4cos(2X)[/tex]

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- Thread starter truewt
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- #1

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[tex] 4cosec(X) - 3sec(X) = 4cos(2X)[/tex]

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- #2

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- #3

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Yup, to both queries.

I tried many approaches but can't solve :(

I tried many approaches but can't solve :(

- #4

rock.freak667

Homework Helper

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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.

EDIT: wait...that might get complicated.

I doubt it has "nice" solutions though.

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- #5

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

- #6

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what I will end up with is [tex] 4cos(X) - 3sin(X) = sin(4X) [/tex]

- #7

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- #8

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

- #9

tiny-tim

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Hi truewt!what I will end up with is [tex] 4cos(X) - 3sin(X) = sin(4X) [/tex]

Hint: write it [tex] \frac{4}{5}cos(X) - \frac{3}{5}sin(X) = \frac{1}{5}sin(4X) [/tex]

Then let sin(Y) = 4/5.

Does that help?

- #10

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The right side becomes:

Put them together and you get

- #11

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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

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I guess you can change the equation to,

[tex] 4cosec(X) - 3sec(X) = 4cos(2X)[/tex]

[tex] (1/2)*cos^{-1}((4cosec(X) - 3sec(X))/4) = X[/tex]

So let

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

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

- #13

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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.

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 get4.55 = 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?

- #14

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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.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.

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

Avodyne

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