Question

cos4x+2cos^2x=0 fastt plss

Answer 1

cos4x = 2cos^2(2x) - 1.
2cos^2x = cos2x + 1.
Now,
cos4x + cos^2x = 0.
2cos^2(2x) - 1 + cos2x + 1 = 0.
cos2x ( 2cos2x + 1 ) = 0.
cos2x = 0 or cos2x = -1/2.
2x = 2n (pi) (+-) (pi)/2 or 2x = 2n (pi) (+-) (pi)/3.
x = n (pi) (+-) (pi)/4 or x = n (pi) (+-) (pi)/6.