Question

Tìm x : Sin (2x-\(\dfrac{1}{3}\))=\(\dfrac{1}{5}\)

Answer 1

\(Sin\left(2x-\dfrac{1}{3}\right)=\dfrac{1}{5}\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-\dfrac{1}{3}=arcsin\left(\dfrac{1}{5}\right)+k2\pi\\2x-\dfrac{1}{3}=\pi-arcsin\left(\dfrac{1}{5}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=arcsin\left(\dfrac{1}{5}\right)+k2\pi+\dfrac{1}{3}\\2x=\pi-arcsin\left(\dfrac{1}{5}\right)+k2\pi+\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{arcsin\left(\dfrac{1}{5}\right)+k2\pi+\dfrac{1}{3}}{2}\\x=\dfrac{\pi-arcsin\left(\dfrac{1}{5}\right)+k2\pi+\dfrac{1}{3}}{2}\end{matrix}\right.\)

Answer 2

=>2x-1/3=arcsin(1/5)+k2pi hoặc 2x-1/3=pi-arcsin(1/5)+k2pi

=>x=1/2(1/3+arcsin(1/5)+k2pi) hoặc x=1/2(1/3+pi-arcsin(1/5)+k2pi)