# Stuck at solving this line

1. Dec 11, 2006

### Dr Game

I've been trying to invent a formula for solving certain sides and angels on objects... and I've tried many different ways to get the formula for it, I think I might have it, but theres this one line I can't figure out how to solve... help would be appreciated.

-4 ln | (csc53 + cot53) = -3 ln | (cscx + cot x)

-2.78 = -3 ln | (cscx + cot x)

-2.78 / 3 = -ln ( (1/sinx) + (1/tanx))

then... how can I get x from that....

2. Dec 11, 2006

### cristo

Staff Emeritus
Well, this is how I'd do it.

e^(2.78/3)=1/sinx+cosx/sinx.

Asinx=1+cosx, where A is the constant on the LHS.

Asinx-cosx=1. Then just use a computer to solve this?

3. Dec 11, 2006

### arildno

Well, remove the minus signs, and get:
$$\ln(\frac{1+\cos(x)}{\sin(x)})=\frac{2.78}{3}$$
whereby:
$$\frac{1+\cos(x)}{\sin(x)}=e^{\frac{2.78}{3}}$$
Agreed so far?

4. Dec 11, 2006

### Dr Game

then is not 1+cos an identity of something? hmm

Last edited: Dec 11, 2006
5. Dec 11, 2006

### Dr Game

telll me if this was an illigal move...

1+cos(x) = 2.526 / sin (x)

1 = 2.526 tan(x)

1/2.526 = tan(x)

x = 21.3 degrees

6. Dec 11, 2006

### cristo

Staff Emeritus
I presume your first line was meant to read 1+cosx=2.526 sinx (typo?)

However, yes that is an "illegal move". You have divided both sides by cosx, but have missed off the term 1/cosx from the LHS

7. Dec 11, 2006

### Dr Game

ya, that was a typo.. and I realized that after I submited that