sin(π–2x)=sin2x
sin4x=2*sin2x*cos2x
2*sin2x*cos2x+sin^22x=0
sin2x*(2cos2x+sin2x)=0
sin2x=0 или 2cos2x+sin2x=0
2x=πk, k ∈ Z
[m]x=\frac{\pi}{2}[/m]k, k ∈ Z
или
sin2x=-2cos2x
Делим на cos2x
tg2x=-2
2x=arctg(-2)+πn, n ∈ Z
[m]x=\frac{1}{2}arctg(-2)+\frac{\pi}{2} n, n \in Z[/m]
О т в е т.
[m]x=\frac{\pi}{2}[/m]k, k ∈ Z
[m]x=\frac{1}{2}arctg(-2)+\frac{\pi}{2} n, n \in Z[/m]