Question

Неравенство sin^2(x)-cos^2(2x)>0. Оригинал во вложении

Answer 1

sin²x-cos²2x>0
(1-cos2x)/2-(1+cos4x)/2>0
1-cos2x-1-cos4x>0
cos4x+cos2x<0
2cos²2x+cos2x-1<0
cos2x=a
2a²+a-1<0
D=1+8=9
a1=(-1-3)/4=-1 U a2=(-1+3)/4=1/2
-1<a<1/2⇒-1<cos2x<1/2
-π+2πn<2x<-π/3 U π/3+2πn<2x<π+2πn
x∈(-π/2+πn;-π/6+πn) U (π/6+πn;π/2+πn)