# Both Sin[x]==0 && Cos[x]==0 as a solution

If I do

FullSimplify[Reduce[Sin[p1] == 0 && Cos[p1] == 0, Reals]]

I get

Cos[p1] == 0 && Sin[p1] == 0

while I would expect False. why? is there a way to have mathematica compute that both sin and cos cannot be 0?

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Reduce[Sin[p1] == 0 && Cos[p1] == 0, p1] returns the expected result. – J. M. Nov 10 '12 at 17:03
Domain specification Reals is unnecessary. – Artes Nov 10 '12 at 17:12
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Thank you @J.M., If you write it as an answer I will mark it as definitive – Fabio Dalla Libera Nov 10 '12 at 17:33