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 = $\frac{n\pi}{2}$ , 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 $\frac{\pi}{4}$ We have, tan2x = –1 Thus tan 2x = tan $\left(\pi - \frac{\pi}{4}\right)$ ⇒ tan 2x = tan $\left(\frac{3\pi}{4}\right)$ ⇒ 2x = nπ + $\frac{3\pi}{4}$ , n ∊ Z Since if tan x = tan α , then x = nπ + α ,n ∊ Z ⇒ x = $\frac{n\pi}{2} + \frac{3\pi}{8}$ , n ∊ Z Hence x = $\frac{n\pi}{2}$ or $\frac{n\pi}{2}+\frac{3\pi}{8}$ , n ∊ Z