Question

solve the eq. fast plz​

Answer 1

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$= > { \cos }^{2} [ \frac{ \pi}{4}(sin \: x + \sqrt{2} { \cos }^{2}) ] - { \tan }^{2} [x + \frac{ \pi }{4} { \tan }^{2} x] = 1........(i)$

$= > { \sin }^{2} [ \frac{ \pi}{4}(sin \: x + \sqrt{2} { \cos }^{2}) ] + { \tan }^{2} [x + \frac{ \pi }{4} { \tan }^{2} x] = 0........(ii)$

IT IS POSSIBLE ONLY WHEN,

$= > { \sin }^{2} [ \frac{ \pi}{4}(sin \: x + \sqrt{2} { \cos }^{2}x) ] = 0 \: \: and \: \: { \tan }^{2} [x + \frac{ \pi }{4} { \tan }^{2} x] = 0$

$\therefore \frac{ \pi}{4}(sin \: x + \sqrt{2} { \cos }^{2}x) = nπ, nϵI$

$or \: \sin x + \sqrt{2} { \cos}^{2} x = 4x$

$\small{ This \: equation \: has \: \: solution \: only \: for} n=0.$

Thus,

$\sin x + \sqrt{2} { \cos }^{2} x = 0$

$\sqrt{2} { \sin }^{2} x - \sin x - \sqrt{2} = 0$

$\therefore \sin x = \frac{ - 1}{ \sqrt{2} } = > x = 2k \pi - \frac{ \pi}{4} ,kϵZ$

Also these values of x satisfy Eq. (ii); therefor, the general solution of given equation is given by

$\boxed{ x = 2k \pi - \frac{ \pi}{4} , kϵz}$

Answer 2

Answer:

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Step-by-step explanation:

$x = 2k\pi$