Ответ:
[2Пk;П(2k+1)], k Z
Объяснение:
0<=2sinx<=2
0<=sinx<=1
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Ответ:
[2Пk;П(2k+1)], k Z
Объяснение:
0<=2sinx<=2
0<=sinx<=1
[2Пk;П(2k+1)], k Z