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roselinBest ResponseYou've already chosen the best response.0
sorry,sorry. its cos x
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Hmm do you remember what the graph for cosine looks like?
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
I'm pretty sure cos(x) = x only has one solution
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
http://screencast.com/t/sLkE6Ylvzpk
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
dw:1351373928927:dw This is kind of the idea rose :D At this particular point, the x value and the cosx value are the same.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
So in your picture, x=pi/2 but cosx = 0 Hmm that point doesn't work so well :c
 one year ago

roselinBest ResponseYou've already chosen the best response.0
okay, i will consider the picture that you showed
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Yah I can't think of the proper way to explain it :D It probably has something to do with the fact that cosine oscillates back and forth between 1 and 1, and since it is continuous it would have to have a solution :d i dunno... just think about it i guess :3 heh
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
Since the function y = x has a domain of all real numbers... as is cos(x), there must be a point of intersection somewhere
 one year ago

roselinBest ResponseYou've already chosen the best response.0
So which one should I consider now? i have two answers here
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
Combine them... you could say.. Since both y=x and the cosine function have a domain of all real numbers and both are continuous at least one point of intersection is definite.
 one year ago

roselinBest ResponseYou've already chosen the best response.0
oh okay, thank you so much guys
 one year ago

ZarkonBest ResponseYou've already chosen the best response.2
use the intermediate value theorem
 one year ago

latremese40Best ResponseYou've already chosen the best response.0
roelin i think you did it right
 one year ago
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