Sin^2x/2-cos^2x/2=cos2x

Sin² (x/2) - cos² (x/2) = cos2x

(1-cosx) / 2 - (1+cosx) / 2=2cos²x-1

1-cosx-1-cosx=4cos²x-2

4cos²x+2cosx-2=0

2cos²x+cosx-1=0

Пусть cosx=t (|t|≤1)

2t²+t-1=0

D=1+8=9; √D=3

t1 = (-1+3) / 4=1/2

t2 = (-1-3) / 4=-1

замена cosx=1/2

x=±π/3+2πn, n € Z

cosx=-1

x=±π+2πn, n € Z