0
$\begingroup$

What can I do to solve this task?

Solve[Power[x, 2] + Power[y, 2] + 1 == cos[Power[x, 2]*Power[y, 2]], {x, 
  y}]
$\endgroup$
0
3
$\begingroup$
Reduce[Power[x, 2] + Power[y, 2] + 1 == 
  Cos[Power[x, 2]*Power[y, 2]], {x, y}, Reals]

x==0&&y==0

$\endgroup$
2
  • $\begingroup$ thank you! Cheers $\endgroup$ – Queno May 28 at 7:02
  • 1
    $\begingroup$ Also, Solve[Power[x, 2] + Power[y, 2] + 1 == Cos[Power[x, 2]*Power[y, 2]], {x, y}, Reals, Method -> Reduce] $\endgroup$ – Bob Hanlon May 29 at 15:39

Not the answer you're looking for? Browse other questions tagged or ask your own question.