Question

Tan^2(2x-pi/4)=3

Tan^2x+cot^2x=2

Answer 1

ĐKXĐ: ...

a.

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

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

\(\Leftrightarrow...\)

b.

\(\Leftrightarrow tan^2x+cot^2x-2=0\)

\(\Leftrightarrow\left(tanx-cotx\right)^2=0\)

\(\Leftrightarrow tanx=cotx=tan\left(\frac{\pi}{2}-x\right)\)

\(\Leftrightarrow x=\frac{\pi}{2}-x+k\pi\)

\(\Leftrightarrow...\)