Question -
Answer -
It is given that
sec2 2x= 1 – tan 2x
We can write it as
1 + tan2 2x= 1 – tan 2x
tan2 2x+ tan 2x = 0
Taking common terms
tan 2x (tan 2x + 1) =0
Here
tan 2x = 0 or tan 2x +1 = 0
If tan 2x = 0
tan 2x = tan 0
We get
2x = nπ + 0, where n ∈ Z
x = nπ/2, where n ∈ Z
tan 2x + 1 = 0
We can write it as
tan 2x = – 1
So we get
Here
2x = nπ + 3π/4, wheren ∈ Z
x = nπ/2 + 3π/8, wheren ∈ Z
Hence, the generalsolution is nπ/2 or nπ/2 + 3π/8, n ∈ Z.