3(1 - cos^2x) + cosx - 1 = 0
3 - 3cos^2x + cosx - 1 = 0
- 3cos^2x + cosx + 2 = 0 //:(-1)
3cos^2x - cosx - 2 = 0
Пусть cosx = t , причём t ∈ [ - 1; 1]
Тогда 3t^2 - t - 2 = 0
D = 1 + 4*3*2 = 25 = 5^2
t1 = ( 1 + 5)/6 = 1;
t2 = ( 1 - 5)/6 = - 2/3
cosx = 1
x = 2pik, k ∈ Z
cosx = - 2/3
x = ± arccos (-2/3) + 2pik
x = ± (pi - arccos(2/3)) + 2pik, k ∈ Z