Odgovor:
# X = 2npi + - (2pi) / 3 #
Obrazloženje:
# Rarrcos2x + 5cosx + 3 = 0 #
# rarr2cos ^ 2x-1 + 3 + 5cosx = 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 #
Ili, # 2cosx + 1 = 0 #
# Rarrcosx = -1 / 2 = cos ((2pi) / 3) #
# Rarrx = 2npi + - (2pi) / 3 # gdje # NrarrZ #
Ili, # Cosx + 2 = 0 #
# Rarrcosx = -2 # što je neprihvatljivo.
Dakle, opće rješenje je # X = 2npi + - (2pi) / 3 #.
Odgovor:
# Theta = 2kpi + - (2pi) / 3 kinZ #
Obrazloženje:
# Cos2theta + 5costheta + 3 = 0 #
#:. 2cos ^ 2 theta-1 + 3 + 5costheta = 0 #
#:. 2cos ^ 2 theta + 5costheta + 2 = 0 #
#:. 2cos ^ 2 theta + 4costheta + costheta + 2 = 0 #
#:. 2costheta (costheta + 2) + 1 (+ costheta 2) = 0 #
#:. (costheta + 2) (2costheta + 1) = 0 #
# => costheta = -2! u -1,1 ili costheta = -1 / 2 #
# => Costheta = cos (pi-pi / 3) = cos ((2pi) / 3) #
# Theta = 2kpi + - (2pi) / 3 kinZ #
Odgovor:
Koristiti # cos2theta = 2 (costheta) ^ 2-1 # i opće rješenje #costheta = cosalpha # je # Theta = 2npi + -alfa #; # N Z #
Obrazloženje:
# Cos2theta + 5costheta + 3 #
# = 2 (costheta) ^ 2-1 + 5kosteta + 3 #
# = 2 (costheta) ^ 2 + 5costheta + 2 #
#rArr (costheta + 1/2), (costheta + 2) = 0 #
Ovdje #costheta = -2 # nije moguće
Dakle, nalazimo samo opća rješenja # Costheta = -1/2 #
# RArrcostheta = (2pi) / 3 #
#:. theta = 2npi + - (2pi) / 3; n Z #
