Question

giải pt sau: sin (\(\dfrac{\pi}{4}\)-x) = cos 2x 

 

Answer 1

\(\Leftrightarrow\sin\left(\dfrac{\Pi}{4}-x\right)=\sin\left(\dfrac{\Pi}{2}-2x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\Pi}{2}-2x=\dfrac{\Pi}{4}-x+k2\Pi\\\dfrac{\Pi}{2}-2x=\dfrac{3}{4}\Pi+x+k2\Pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=\dfrac{-\Pi}{4}+k2\Pi\\-3x=\dfrac{1}{4}\Pi+k2\Pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{4}-k2\Pi\\x=\dfrac{-1}{12}\Pi-\dfrac{2}{3}k\Pi\end{matrix}\right.\)