Question

find the general solution of sec^2 2x = 1- tan2x

Answer 1

$\underline{\large{\sf Answer:}}$

Here, we have given equation,

$\sf sec^2 2x = 1 - tan2x$

we know, 1 + tan²Φ = sec²Φ

∴ $\sf (1 + tan^2 2x) = (1 - tan2x)$

∴ $\sf (1- 1 + tan^22x + tan2x) =0$

$\implies \sf (tan^22x + tan2x) = 0$

$\implies \sf tan 2x(tan 2x + 1)=0$

$\implies \sf tan 2x = 0\:OR\:tan 2x + 1 = 0$

∴ $\sf tan2x = 0\:OR\:tan2x = -1$

$\sf tan 2x = -1 = -tan(\frac{\pi}{4})$

$\sf =-(- tan(\pi - \frac{\pi}{4}))= tan (\frac{3\pi}{4})$

We know,

tan Φ = 0 implies Φ = nπ and ,

tanΦ = tanα implies Φ = nπ + α ,n€Z.

Here ,

$\sf tan2x = 0\:OR\:tan2x=tan(\frac{3\pi}{4})$

∴ $\sf 2x = n\pi\:OR\:2x=m\pi+(\frac{3\pi}{4})$

$\implies \sf x = (\frac{n\pi}{2})\:OR\:x=(\frac{m\pi}{2})+(\frac{3\pi}{8})$

therefore, the general solutions of the given equation are $\sf x = (\frac{n\pi}{2})\:OR\:x=(\frac{m\pi}{2})+(\frac{3\pi}{8})$,here n, m € Z