Sagot:
# x = 2npi + - (2pi) / 3 #
Paliwanag:
# 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 #
Alinman, # 2cosx + 1 = 0 #
# rarrcosx = -1 / 2 = cos ((2pi) / 3) #
# rarrx = 2npi + - (2pi) / 3 # kung saan # nrarrZ #
O kaya, # cosx + 2 = 0 #
# rarrcosx = -2 # na kung saan ay hindi katanggap-tanggap.
Kaya, ang pangkalahatang solusyon ay # x = 2npi + - (2pi) / 3 #.
Sagot:
# theta = 2kpi + - (2pi) / 3, kinZ #
Paliwanag:
# cos2theta + 5costheta + 3 = 0 #
#: 2cos ^ 2theta-1 + 5costheta + 3 = 0 #
#: 2cos ^ 2theta + 5costheta + 2 = 0 #
#:. 2cos ^ 2theta + 4costheta + costheta + 2 = 0 #
#: 2costheta (costheta + 2) +1 (costheta + 2) = 0 #
#:. (costheta + 2) (2costheta + 1) = 0 #
# => costheta = -2! sa -1,1, o costheta = -1 / 2 #
# => costheta = cos (pi-pi / 3) = cos ((2pi) / 3) #
# theta = 2kpi + - (2pi) / 3, kinZ #
Sagot:
Gamitin # cos2theta = 2 (costheta) ^ 2-1 # at ang pangkalahatang solusyon ng #costheta = cosalpha # ay # theta = 2npi + -alpha #; # n Z #
Paliwanag:
# cos2theta + 5costheta + 3 #
# = 2 (costheta) ^ 2-1 + 5costheta + 3 #
# = 2 (costheta) ^ 2 + 5costheta + 2 #
#rArr (costheta + 1/2) (costheta + 2) = 0 #
Dito #costheta = -2 # ay hindi posible
Kaya, nakita lamang natin ang pangkalahatang mga solusyon ng # costheta = -1 / 2 #
# rArrcostheta = (2pi) / 3 #
#: theta = 2npi + - (2pi) / 3; n Z #
