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(A) Given equation: $\sec^{2} 2x = 1 - 	an 2x$Using the identity $\sec^{2} 	heta = 1 + 	an^{2} 	heta$,we get:$1 + 	an^{2} 2x = 1 - 	an 2x$$	an^

Vedclass · Accepted Answer

$x = \frac{n\pi}{2}$ or $x = \frac{n\pi}{2} + \frac{3\pi}{8}, n \in \mathbb{Z}$

Vedclass · Answer

(A) Given equation: $\sec^{2} 2x = 1 - 	an 2x$Using the identity $\sec^{2} 	heta = 1 + 	an^{2} 	heta$,we get:$1 + 	an^{2} 2x = 1 - 	an 2x$$	an^{2} 2x + 	an 2x = 0$$	an 2x(	an 2x + 1) = 0$This gives two cases:Case $1$: $	an 2x = 0$$2x = n\pi$,where $n \in \mathbb{Z}$$x = \frac{n\pi}{2}, n \in \mathbb{Z}$Case $2$: $	an 2x = -1$$	an 2x = -	an \frac{\pi}{4} = 	an(\pi - \frac{\pi}{4}) = 	an \frac{3\pi}{4}$$2x = n\pi + \frac{3\pi}{4}$,where $n \in \mathbb{Z}$$x = \frac{n\pi}{2} + \frac{3\pi}{8}, n \in \mathbb{Z}$Thus,the general solution is $x = \frac{n\pi}{2}$ or $x = \frac{n\pi}{2} + \frac{3\pi}{8}, n \in \mathbb{Z}$.

Find the general solution of the equation $\sec^{2} 2x = 1 - \tan 2x$.

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