Solving for unknown in arccos equation help?

[calky117]

I need help with the following, is it possible to get the below equation into a form like:

THETA = 30a

or somehow get the 'a' on the left hand side?

Equation:

THETA = arccos(1 - 0.005a)Those aren't the exact numbers/arrangement I'm working with but any help would be greatly appreciated.

 

[eumyang]

I need help with the following, is it possible to get the below equation into a form like:

THETA = 30a

or somehow get the 'a' on the left hand side?

Equation:

THETA = arccos(1 - 0.005a)


Those aren't the exact numbers/arrangement I'm working with but any help would be greatly appreciated.

It would be better if you posted the original problem instead of using numbers and/or an arrangement that isn't exact. Are you trying to solve for a? If so, as long as you're mindful of the domain/range restrictions, why can't you take the cosine of both sides?

 

[calky117]

Sorry it is a bit hard to post the entire problem as this is only part of quite a large engineering question.

From the above the unknown is 'x' which is what I want to solve for (again I have skipped out a lot of the working that is irrelevant to you).

If I substitute in THETA to the area 'A' equation then put that into the equation on the bottom I get a huge awful looking equation, only the one unknown but I just don't know how to solve for it. Any guidance please?

 

[Mark44]

You're starting with cos(\theta/2) = (150 - .77x)/150.
You solved for \theta/2, but didn't solve for x.

To isolate x (i.e., solve for x), multiply both sides of the original equation by 150. Then add -150 to both sides. Finally, divide both sides by -.77 to get x all by itself.