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When I plug in an equation involving ArcSin to SolveValues, I get a result involving Root and a pure function, with the answer embeded.

Here is a simplified version of the equation:

SolveValues[1 == a ArcSin[a], a, Assumptions -> 0 <= a <= 1 ]

Output:

{
Root[{
   -1 + ArcSin[#] # &, 
   0.89753946128048718439307346196209434156`20.300813366021398
    }]
}

The long string of digits is the answer I want, but why is it given inside this expression, and how to I get it out?

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  • $\begingroup$ Try this: FindRoot[1 == a ArcSin[a], {a, 0.5}]. $\endgroup$ Oct 1, 2022 at 15:12
  • 2
    $\begingroup$ The Root expression is the exact solution, i.e., an exact number. Like any other exact number (e.g., Sin[2]), use N to convert it to an approximate value. $\endgroup$
    – Bob Hanlon
    Oct 1, 2022 at 15:14
  • $\begingroup$ I know the first part, I was including it in an answer to my own post! The second part is what I didn't know $\endgroup$
    – Cat Bisque
    Oct 1, 2022 at 15:15
  • 2
    $\begingroup$ Related: mathematica.stackexchange.com/questions/13767/… $\endgroup$
    – Michael E2
    Oct 1, 2022 at 15:17
  • $\begingroup$ Thanks for that reference, @MichaelE2. $\endgroup$ Oct 1, 2022 at 15:20

2 Answers 2

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Try this:

NSolveValues[1 == a ArcSin[a] && 0 <= a <= 1, a]
(*{0.8975394612804872`}*)

Or, as @Bob Hanlon points out:

SolveValues[1 == a ArcSin[a], a, Assumptions -> 0 <= a <= 1] // N
(*{0.8975394612804872`}*)
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From the documentation, the Root function "Root[{f,c}] represents the exact root of the general equation f[x] = 0 near x = c."

Since 0.89753946128048718439307346196209434156`20.300813366021398 is not an exact answer,

{
Root[{
   -1 + ArcSin[#] # &, 
   0.89753946128048718439307346196209434156`20.300813366021398
    }]
}

is given to specify the exact answer, which occurs near 0.8975...

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