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OpenStudy (anonymous):
Yeah, there is.
OpenStudy (anonymous):
Can you explain?
sam (.sam.):
Newton's method?
OpenStudy (anonymous):
You basically have to find the root of \[
\cos(x)-x=0
\]
OpenStudy (anonymous):
Think about it this way...
As \(x\) goes from \(0\) to \(\pi\), \(\cos x\) goes from \(1\) to \(0\).
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OpenStudy (anonymous):
And this happens in a one-to-one manner.
OpenStudy (anonymous):
Oh ok that makes sense. Thank you so much this really helps
OpenStudy (anonymous):
So they have to meet at some point.
OpenStudy (anonymous):
Should maybe say as \(x\) goes from \(0\) to \(\pi/2\)
OpenStudy (anonymous):
Although this wouldn't be very easy to solve algebraically, but one thing you know for sure is that this will have 1 solution since the line y = x and the curve y = cos(x) have to intersect at some point.
@skater93
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OpenStudy (anonymous):
It has 2 solutions, because the negative case is symmetrically identical.