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