Hoe los je cos 2theta + 5 cos theta + 3 = 0 op?

Antwoord:

# X = 2npi + - (2pi) / 3 #

# 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 #

Een van beide, # 2cosx + 1 = 0 #

# Rarrcosx = -1/2 = cos ((2pi) / 3) #

# Rarrx = 2npi + - (2pi) / 3 # waar # NrarrZ #

Of, # Cosx + 2 = 0 #

# Rarrcosx = -2 # wat onaanvaardbaar is.

Dus de algemene oplossing is # X = 2npi + - (2pi) / 3 #.

# Theta = 2kpi + - (2pi) / 3, Kinz #

# 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! in -1,1, of costheta = -1 / 2 #

# => Costheta = cos (pi-pi / 3) = cos ((2pi) / 3) #

# Theta = 2kpi + - (2pi) / 3, Kinz #

Gebruik # cos2theta = 2 (costheta) ^ 2-1 # en de algemene oplossing van #costheta = cosalpha # is # Theta = 2npi + alfa #; # N Z #

# Cos2theta + 5costheta + 3 #

# = 2 (costheta) ^ 2-1 + 5costheta + 3 #

# = 2 (costheta) ^ 2 + 5costheta + 2 #

#rArr (costheta + 1/2) (costheta + 2) = 0 #

Hier #costheta = -2 # is niet mogelijk

Dus we vinden alleen de algemene oplossingen van # Costheta = -1/2 #

# RArrcostheta = (2pi) / 3 #

#:. theta = 2npi + - (2pi) / 3; n Z #