Question

dùng công thức hạ bậc giải pt sau \(tan^2x\left(2x-\dfrac{\pi}{3}\right)=2\)

Answer 1

=>\(\left\{{}\begin{matrix}tan\left(2x-\dfrac{pi}{3}\right)=\sqrt{2}\\tan\left(2x-\dfrac{pi}{3}\right)=-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-\dfrac{pi}{3}=arctan\left(\sqrt{2}\right)+kpi\\2x-\dfrac{pi}{3}=arctan\left(-\sqrt{2}\right)+kpi\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\left(\dfrac{pi}{3}+arctan\left(\sqrt{2}\right)+kpi\right)\\x=\dfrac{1}{2}\left(\dfrac{pi}{3}+arctan\left(-\sqrt{2}\right)+kpi\right)\end{matrix}\right.\)