Given that,
cos4x=cos2x cos4x−cos2x=0 ( ∵cosx−cosy=−2sin x+y 2 sin x−y 2 ) −2sin3xsinx=0 sin3xsinx=0
So, either sin3x=0 or sinx=0.
General solution for sin3x=0,
3x=nπ x= nπ 3
Where, n∈z
General solution for sinx=0,
x=nπ
Where, n∈z
Thus, the general solutions are,
For sin3x=0, x= nπ 3
For sinx=0, x=nπ where, n∈z