Kako rješavate cos 2x-sin ^ 2 (x / 2) + 3/4 = 0?

Odgovor:

# Cosx = 1/2 # i # Cosx = -3/4 #

Korak 1:

# Cos2x-Sin ^ 2 (x / 2) + 3/4 = 0 #

Koristiti # Cos2x = cos ^ 2 x-sin ^ 2 x #

Korak 2:

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

Koristiti # Grijeh ^ 2x + cos ^ 2 x = 1 #

Korak 3:

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

Koristiti # Cosx = 1-2sin ^ 2 (x / 2) * (Formula dvostrukog kuta).

Korak 4:

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

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

Pomnožite sa 4 da biste dobili

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

Korak 5: Riješite kvadratnu jednadžbu

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

# Cosx = 1/2 # i # Cosx = -3/4 #