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

Odgovor:

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

Korak 1:

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

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

2. korak:

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

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

3. korak:

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

Uporaba # cosx = 1-2sin ^ 2 (x / 2) # (Formula z dvojnim kotom).

4. korak:

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

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

Pomnožite s 4, da dobite

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

5. korak: Rešite kvadratno enačbo

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

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