funkcje koko: jak obliczyć miejsca zerowe funkcji? cos (−4x−2π) i −ctg|3πx|)

Mila: cos (−4x−2π) =0⇔

	π
−4x−2π=k		i keC
	2

	π
−4x=2π+k
	2

x=... ctg(3πx)=0 i x≠kπ odczytaj z wykresu

Mila: cos (−4x−2π) =0⇔

	π
−4x−2π=k		i keC
	2

	π
−4x=2π+k
	2

x=... ctg(3πx)=0 i x≠kπ odczytaj z wykresu

Mila: cos (−4x−2π) =0⇔

	π
−4x−2π=k		i keC
	2

	π
−4x=2π+k
	2

x=... ctg(3πx)=0 i x≠kπ odczytaj z wykresu

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Basia: Milu cosx = 0 ⇔ x = ^π₂+kπ można napisać, że ⇔ x=(2k+1)*^π₂ ale nie x=k*^π₂ bo wtedy miałabyś dla k=0 cos0=0 co oczywiście jest nieprawdą

Mila: Koko ma być tak. cos (−4x−2π) =0⇔

	π
−4x−2π=		+kπ
	2

Mila: Basiu, nie wiem dlaczego tak głupio napisałam, sprawdziłam swoje notatki i wszędzie mam dobrze. Chyba zbyt późno to pisałam. Dzięki za ostrzeżenie. Myślę jednak, że nasz uczeń wykryłby szybko tę usterkę.