Troubleshooting a Formula for Solving Sides and Angles: Help Needed!"

[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 there's 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...

 

[cristo]

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?

 

[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?

 

[Dr Game]

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

 

[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

 

[cristo]

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

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

 

[Dr Game]

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