Paano mo malulutas ang cos 2x-sin ^ 2 (x / 2) + 3/4 = 0?

Sagot:

# Cosx = 1/2 # at # cosx = -3 / 4 #

Hakbang 1:

# cos2x-Sin ^ 2 (x / 2) + 3/4 = 0 #

Gamitin # cos2x = cos ^ 2x-sin ^ 2x #

Hakbang 2:

# cos ^ 2x-sin ^ 2x-sin ^ 2 (x / 2) + 3/4 = 0 #

Gamitin # sin ^ 2x + cos ^ 2x = 1 #

Hakbang3:

# 2cos ^ 2x-1-sin ^ 2 (x / 2) + 3/4 = 0 #

Gamitin # cosx = 1-2sin ^ 2 (x / 2) # (Double formula ng anggulo).

Hakbang 4:

# 2cos ^ 2x-1-1 / 2 + 1 / 2cosx + 3/4 = 0 #

# 2cos ^ 2x + 2cosx-3 = 0 #

Multiply ng 4 upang makuha

# 8cos ^ x + 2cosx-3 = 0 #

Hakbang 5: Lutasin ang parisukat na equation upang makuha

# (2cos-1) (4cosx + 3) = 0 #

# cosx = 1/2 # at # cosx = -3 / 4 #