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 =\( \dfrac{n\pi}{2}\), where n ∈ Z
tan 2x + 1 = 0
We can write it as
tan 2x = – 1
So we get
\(- tan\dfrac{\pi}{4}= tan(\pi-\dfrac{\pi}{4})\)
Here
2x = nπ + \( \dfrac{3\pi}{4}\), where n ∈ Z
x = \( \dfrac{n\pi}{2}+ \dfrac{3\pi}{8}\) \(\), where n ∈ Z
Hence, the general solution is \( \dfrac{n\pi}{2}\)or \( \dfrac{n\pi}{2}\) +\( \dfrac{3\pi}{8}\), n ∈ Z.
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