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Zad. 1)
sin(πcosx)=0
Niech πcosx = t
wtedy :
sint = 0
t = 2kπ lub t = π + 2kπ
ale t = πcosx,
więc:
πcosx= 2kπ/:π lub πcosx = π + 2kπ /:π
cosx = 2k lub cosx = 1 + 2k
x = 2k + 2kπ lub x = - 2k + 2kπ
Zad. 2)