Question

\(tgx+cotgx=2\\ 3tg^4x+2tg^2x-1=0\)

Answer 1

a: \(\Leftrightarrow\dfrac{2}{\sin2x}=2\)

\(\Leftrightarrow\sin2x=1\)

\(\Leftrightarrow2x=\dfrac{\Pi}{2}+k2\Pi\)

hay 

b: \(\Leftrightarrow3\cdot tan^4x+3tan^2x-tan^2x-1=0\)

\(\Leftrightarrow3tan^2x-1=0\)

\(\Leftrightarrow tan^2x=\dfrac{1}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=arctan\left(\dfrac{1}{\sqrt{3}}\right)+k\Pi=\dfrac{\Pi}{6}+k\Pi\\x=-\dfrac{\Pi}{6}+k\Pi\end{matrix}\right.\)