Question

\(\tan^2\left(2x-\frac{\pi}{4}\right)=3\)

Answer 1

ĐKXĐ: ...

\(\Leftrightarrow\left[{}\begin{matrix}tan\left(2x-\frac{\pi}{4}\right)=\sqrt{3}\\tan\left(2x-\frac{\pi}{4}\right)=-\sqrt{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{4}=\frac{\pi}{3}+k\pi\\2x-\frac{\pi}{4}=-\frac{\pi}{3}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7\pi}{24}+\frac{k\pi}{2}\\x=-\frac{\pi}{24}+\frac{k\pi}{2}\end{matrix}\right.\)