Solving A Trigonometric Equation | Problem 354

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[@scottleung9587]

Got it!

[@Blaqjaqshellaq]

The Taylor expansion of Arccos(i) is pi/2 - i + i/6 - i*3/40 + i*5/112 - i*7/2304 + ...

[@marvinco33]

using the sine, arctangent ([+-sqrt2]/i) gives the same decimal approx result

[@phill3986]

😃✌️👍👍✌️😃

[@trojanleo123]

z = π/2 - i*ln(√2 ± 1)

[@quneptune]

I wonder if something like z-sin(z)=1/4 would be possible with the complex definition of sine