The given equation can be re-written as sin
2
4x−2sin4xcos
4
x+cos
2
x=0
Add and subtract cos
8
x
∴(sin4x−cos
4
x)
2
+cos
2
x(1−cos
6
x)=0
Since both the terms are +ive (cos
6
x≤1), above is possible only when each term is zero for the same value of x.
sin4x−cos
4
x=0 .(1)
and cos
2
x(1−cos
6
x)=0 .(2)
From (2) cosx=0 or cos
2
x=1
∵z
3
=1⇒z=1 only
as other values will not be real.
Case I: If cosx=0 i.e., x=(n+
2
1
)π, then from (1)
sin4(n+
2
1
)π+0=0
or sin(4n+2)π=0 which is true.
∴x=(n+
2
1
)π (3)
Case II: When cos
2
x=1 i.e., sinx=0
∴x=rπ then from (1), sin4rπ−1=0 or −1=0 which is not true. Hence the only solution is given by (3).
