Question

Find the general solution of the equation Sec^2 (2x) = 1 - tan2x

Answer 1

sec^2(2x)=1+tg^2(2x)
1+tg^2(2x)=1-tg(2x)
tg^2(2x)+tg(2x)=0
tg(2x)[tg(2x)+1]=0
tg(2x)=0 or tg(2x)+1=0
2x=arctg(0) or 2x=arctg(-1)

Answer 2

\[
\sec^2 (2x) = 1 - \tan2x\\
1 + \tan^2(2x)=1 -\tan(2x)\\
\tan^2(2x)+\tan(2x)=0\\
\tan(2x)( \tan(2x) +1)=0\\
\tan(2x) =0\\
\tan(2 x) =-1

\]

Answer 3

\[
\tan(2x)=0\\
2x =k \pi\\
x= \frac {k \pi}2
\]

Answer 4

\[
\tan(2 x) =-1\\
\tan(2 x) = \tan \left ( - \frac \pi 4 \right)\\
x = -\frac \pi 8 + k \pi

\]