I think this is something ridiculously simple but it just slipped out of my mind.
Say you are given an equation:
0.0036 = p(1-cos(0.6/p))
how would you solve for p (note: cos is measured in radians, not degrees)? The answer given is p = 50, but can someone show me the steps?
Post has been edited 2 time(s), last time on Apr 17 2008, 6:09 am by MillenniumArmy.
None.
The book is wrong, by a factor of ten.
None.
This is all I could get:
0.036 = p(1-cos(0.6/p))
0 = p(1-cos(0.6/p)) - 0.036
0 = p - pcos(0.6/p) - 0.036
cos(nx) = 2 cos([n-1]x)cos(x) - cos([n-2]x)
0.6/p = 3/5/p = 3/5p = 6/10p = 1.2/2p = 3/5 * 1/p
3/5/p = 3/p/5
The book is wrong, by a factor of ten.
LOL oops, i meant 0.0036, not 0.036 mb
None.
Expand to Taylor series and approximate.
None.
I think this is something ridiculously simple but it just slipped out of my mind.
Say you are given an equation:
0.0036 = p(1-cos(0.6/p))
how would you solve for p (note: cos is measured in radians, not degrees)? The answer given is p = 50, but can someone show me the steps?
You cannot really solve that, the solution is probably irrational. But you can approximate it by several methods, including Taylor series.
?????
