1 + tg^2(x) = 2tg^2(x) <=> tg^2(x) = 1 <=>\\
<=> tg(x) = 1;\\
\\
x = \left[\begin{array}{c} arctg(1) + 2\pi k\\ \pi + arctg(1) + 2\pi n\end{array}\right \ k,n \in \mathbb Z\\
\\
x = \left[\begin{array}{c} \frac{\pi}{4} + 2\pi k\\ \frac{5\pi}{4} + 2\pi n\end{array}\right \ k,n \in \mathbb Z
" alt="\frac{1}{cos^2(x)} = 2tg^2(x) <=> 1 + tg^2(x) = 2tg^2(x) <=> tg^2(x) = 1 <=>\\
<=> tg(x) = 1;\\
\\
x = \left[\begin{array}{c} arctg(1) + 2\pi k\\ \pi + arctg(1) + 2\pi n\end{array}\right \ k,n \in \mathbb Z\\
\\
x = \left[\begin{array}{c} \frac{\pi}{4} + 2\pi k\\ \frac{5\pi}{4} + 2\pi n\end{array}\right \ k,n \in \mathbb Z
" align="absmiddle" class="latex-formula">