(sin2x+cos2x)^2=sin^22x+2sin2xcos2x+cos^22x
(sin2x+cos2x)^2=1+2sin2xсos2x.
Уравнение принимает вид:
1–cos2x=1+2sin2xcos2x
2sin2xcos2x+cos2x=0
cos2x·(2sin2x+1)=0
cos2x=0 или sin2x=–1/2
cos2x=0 ⇒2x= (π/2)+πk, k∈Z
sin2x=–1/2⇒
2x=(–π/6)+2πn, n∈Z или 2х=(–5π/6)+2πm, m∈Z
x= (π/4)+(π/2)k, k∈Z
x=(–π/12)+πn, n∈Z или х=(–5π/12)+πm, m∈Z
Указанному промежутку
[3π; 4π]=[36π/12; 48π/12] принадлежат корни:
(π/4)+(π/2)·6=13π/4=39π/12
(π/4)+(π/2)·7=15π/4=45π/12
(–π/12)+4π=47π/12
(–5π/12)+4π=43 π/12
О т в е т.
а)
x= (π/4)+(π/2)k, k∈Z
x=(–π/12)+πn, n∈Z
х=(–5π/12)+πm, m∈Z
б)39π/12; 43 π/12; 45π/12; 47π/12.