Question -
Answer -
It is given that
sec2┬а2x= 1 тАУ tan 2x
We can write it as
1 + tan2┬а2x= 1 тАУ tan 2x
tan2┬а2x+ 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 = n╧А + 0, where n тИИ Z
x = n╧А/2, where n тИИ Z
tan 2x + 1 = 0
We can write it as
tan 2x = тАУ 1
So we get
Here
2x = n╧А + 3╧А/4, wheren тИИ Z
x = n╧А/2 + 3╧А/8, wheren тИИ Z
Hence, the generalsolution is n╧А/2 or n╧А/2 + 3╧А/8, n тИИ Z.