Cevap:
#, X = 2npi + - (2pi) / 3 #
Açıklama:
# Rarrcos2x + 5cosx + 3 = 0 #
# Rarr2cos ^ 2x-1 + 5cosx + 3 = 0 #
# Rarr2cos ^ 2x + 5cosx + 2 = 0 #
# Rarr2cos ^ 2x + 4cosx + cosx + 2 = 0 #
# Rarr2cosx (cosx + 2) + 1 (cosx + 2) = 0 #
#rarr (2cosx + 1) (cosx + 2) = 0 #
Ya # 2cosx + 1 = 0 #
# Rarrcosx = -1/2 = ((2pi) / 3) # cos
# Rarrx = 2npi + - (2pi) / 3 # nerede # NrarrZ #
Veya, # Cosx + 2 = 0 #
# Rarrcosx = -2 # bu kabul edilemez.
Yani, genel çözüm #, X = 2npi + - (2pi) / 3 #.
Cevap:
# Teta = 2kpi + - (2pi) / 3, # Kinz
Açıklama:
# Cos2theta + 5costheta + 3 = 0 #
#:. 2cos ^ 2teta-1 + 5costheta + 3 = 0 #
#:. 2cos ^ 2teta + 5costheta + 2 = 0 #
#:. 2cos ^ 2teta + 4costheta + costheta + 2 = 0 #
#:. 2costheta (costheta + 2) + 1 (costheta + 2) = 0 #
#:. (costheta + 2) (2costheta + 1) = 0 #
# => costheta = -2! -1,1 içinde veya costheta = -1 / 2 #
# => Costheta = (pi-pi / 3) = cos ((2pi) / 3) # cos
# Teta = 2kpi + - (2pi) / 3, # Kinz
Cevap:
kullanım # cos2theta = 2 (costheta) ^ 2-1 # ve genel çözümü #costheta = cosalpha # olduğu # Teta = 2npi + -aIfa #; # N Z #
Açıklama:
# Cos2theta + 5costheta + 3 #
# = 2 (costheta) ^ 2-1 + 5costheta + 3 #
# = 2 (costheta) ^ 2 + 5costheta + 2 #
#rArr (costheta + 1/2) (costheta + 2) = 0 #
İşte #costheta = -2 # imkansız
Yani, biz sadece genel çözümlerini buluruz. # Costheta = -1/2 #
# RArrcostheta = (2pi) / 3 #
#: teta = 2npi + - (2pi) / 3; n Z #
