Question

\(tan^22x-4tan2x=0\)

Phương trình này giải như nào ạ ??

 

Answer 1

ĐK: \(x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

\(tan^22x-4tan2x=0\)

\(\Leftrightarrow tan2x\left(tan2x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tan2x=0\\tan2x=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\2x=arctan4+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\x=\dfrac{1}{2}arctan4+\dfrac{k\pi}{2}\end{matrix}\right.\)

Answer 2

\(tan^22x-4tan2x=0\)

⇒\(tan2x\left(tan-4\right)=0\)