Question

4sin^4x+sin^2 2x=2
Решите пожалуйста

Answer 1

4sin⁴x+sin²2x=2
(2sin²x)²-sin²2x=2
(1-cos2x)²-1+cos²2x-2=0
1-2•cos2x+cos²2x+cos²2x-3=0
cos2x=t
2t²-2t-2=0
t²-t-1=0
D=1+4=5
t1=(1+√5)/2
t2=(1-√5)/2
1)cos2x=(1+√5)/2 ;x€∅
2)cos2x=(1-√5)/2
2x=±arccos((1-√5)/2)+2πk
x=±1/2arccos((1-√5)/2)+πk;k€Z