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Find the general solution of the following equation sec^22x is equal to 1 = tan2x

Q (8)  Find the general solution of the following equation 

   \small \sec^{2}2x = 1 - \tan2x

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We know that 
\sec^{2}x = 1 + \tan^{2}x
So,
       1 + \tan^{2}2x = 1 -\tan2x
       \tan^{2}2x + \tan2x = 0\\ \\ \tan2x(\tan2x+1) = 0
 either
      tan2x = 0             or                     tan2x = -1                                                  (     \tan x = \tan \left ( \pi - \frac{\pi}{4} \right ) = \tan\frac{3\pi}{4} )
            2x = n\pi                       2x=n\pi + \frac{3\pi}{4}
               x=\frac{n\pi}{2}                        x=\frac{n\pi}{2} + \frac{3\pi}{8}
  Where n \epsilon Z

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