Cos^24x+3sin4x-sin^24x = 2

Question

Cos^24x+3sin4x-sin^24x = 2

Answer 1

Cos²4x+3sin4x-sin²4x=2
1-sin²4x+3sin4x-sin²4x-2=0
-2sin²4x+3sin4x-1=0
sin4x=t
-2t²+3t-1=0
D=9-4*(-2)*(-1)=1
t1=-3+1/-4=1/2
t2=-3-1/-4=1
sin4x=1/2
4x=(-1)ⁿп/6+пk,k€z
x=(-1)ⁿп/24+пk/4,k€z
sin4x=1
4x=п/2+2пк,к€z
x=п/8+пк/2,к€z