Why doesn't Mathematica solve x == Cos[x] properly?
Both Solve and NSolve fail with the message:
Solve::nsmet: This system cannot be solved with the methods available to Solve. >>
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3 
2 Answers
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7 In this case you might use:
InverseFunction[Cos] ArcCos
One can see that this is valid only over the interval (-1, 1) which is probably why Solve does not give an answer:
Plot[{Cos@x, ArcCos@x}, {x, -Pi, Pi}, PlotStyle -> Thick] 
A few methods to find the intersection in the illustration:
N @ FindInstance[x == Cos[x], x] N @ Reduce[{x == Cos[x], -1 < x < 1}, x] FindRoot[x == Cos[x], {x, 0}] {{x -> 0.739085}} x == 0.739085 {x -> 0.739085}
I would be remiss not to point out that my plot above is only looking at real values. One can see that as implemented ArcCos does handle the full circle:
Plot[{Cos @ ArcCos @ x, x + 1}, {x, -20, 20}] 
- $\begingroup$ Thx, I just wanted to know why Mathematica didn't know to use arccos... $\endgroup$Tom Wellington– Tom Wellington2012-10-22 07:09:27 +00:00Commented Oct 22, 2012 at 7:09
- $\begingroup$ Of note is that
Reduce[]returns aRoot[]object, which can then be evaluated to arbitrary precision. $\endgroup$J. M.'s missing motivation– J. M.'s missing motivation2012-10-22 07:19:32 +00:00Commented Oct 22, 2012 at 7:19 - 1$\begingroup$ @Mr.Wizard The 0.739... in the argument is not used to compute the value of the root, only to isolate it from any other potential roots. In this sense, it is similar to the index for normal polynomial
Rootobjects, as returned by say, Solve[x^5 -x-1 == 0, x]. Of course, the root can be computed to arbitrary precision using the-Cos[#1] + #1 &. $\endgroup$Mark McClure– Mark McClure2012-10-22 10:32:36 +00:00Commented Oct 22, 2012 at 10:32 - 1$\begingroup$ @TomWellington This answer gives more detailed discussion of the issue mathematica.stackexchange.com/questions/4694/… $\endgroup$Artes– Artes2012-10-22 10:35:43 +00:00Commented Oct 22, 2012 at 10:35
- 1$\begingroup$ As @Mark says, you can treat the
0.739...as a "proximity indicator"; you can thus readRoot[{-Cos[#1] + #1 &, 0.73908513321516064165}]as "the root of the equation $x-\cos\,x=0$ in the vicinity of $0.739085\dots$". You can apply something likeN[#, 100] &to it, and the internal algorithms will then compute the root to higher precision. $\endgroup$J. M.'s missing motivation– J. M.'s missing motivation2012-10-22 11:11:36 +00:00Commented Oct 22, 2012 at 11:11
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2 Another possibility :
FixedPoint[Cos[#] &, 0.5] (* 0.739085 *) answered Oct 22, 2012 at 6:59
b.gates.you.know.what
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- 3$\begingroup$ You don't need
&:FixedPoint[Cos, 0.5]$\endgroup$Mr.Wizard– Mr.Wizard2012-10-22 07:06:18 +00:00Commented Oct 22, 2012 at 7:06 - 1$\begingroup$ @Mr.Wizard Thanks, I was aware of this but thought to make the answer easier to generalize. $\endgroup$b.gates.you.know.what– b.gates.you.know.what2012-10-22 07:11:05 +00:00Commented Oct 22, 2012 at 7:11
NSolve[x==Cos[x],x,Reals]gives{{x -> 0.739085}}. $\endgroup$