cos4x = 6cos^2x - 5;

cos2x = 3 + 7cosx

Question

Алгебра

Янина

23 июня, 01:50

cos4x = 6cos^2x - 5;

cos2x = 3 + 7cosx

+1

Ответы (1)

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Боруля · Answer 1

cos2X = 3 + 7cosX

2cos"2 (X) - 1 = 3 + 7cosX

2cos"2 (X) - 7 cosX - 4 = 0

(cosX - 4) (2cosX + 1) = 0

cosX=4, cosX = - 1/2

X=nevozmojno naiti (tak kak bolwe 1) X = 120

cos4x = 6cos^2x - 5;cos2x = 3 + 7cosx

cos4x = 6cos^2x - 5;

cos2x = 3 + 7cosx