We have, sec2 2x = 1 – tan 2x
⇒ 1 + tan2 2x = 1 – tan 2x ⇒ tan2 2x + tan 2x = 0
⇒ tan 2x (tan 2x + 1) = 0 ⇒ either tan 2x = 0 or, tan 2x + 1 = 0
Now, if tan 2x = 0 ⇒ 2x = nπ ⇒ x =

nπ

2

, n ∊ Z
Since if tan x = 0 , then x = nπ ; n ∊ Z
And, if tan 2x + 1 = 0 ⇒ tan 2x = – 1
A value of x satisfying tanx = 1 is

π

4

We have, tan2x = –1
Thus tan 2x = tan (π−

π

4

) ⇒ tan 2x = tan (

3π

4

)
⇒ 2x = nπ +

3π

4

, n ∊ Z
Since if tan x = tan α , then x = nπ + α ,n ∊ Z
⇒ x =

nπ

2

+

3π

8

, n ∊ Z
Hence x =

nπ

2

or

nπ

2

+

3π

8

, n ∊ Z