\(\sqrt{2}sinx=2sin2x.cos2x-cos2x\)
\(\Leftrightarrow\sqrt{2}sinx=sin4x-cos4x\)
\(\Leftrightarrow\sqrt{2}sinx=\sqrt{2}sin\left(4x-\dfrac{\pi}{4}\right)\)
\(\Leftrightarrow sin\left(4x-\dfrac{\pi}{4}\right)=sinx\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-\dfrac{\pi}{4}=x+k2\pi\\4x-\dfrac{\pi}{4}=\pi-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+\dfrac{k\pi}{3}\\x=\dfrac{\pi}{4}+\dfrac{k2\pi}{5}\end{matrix}\right.\)
Đúng 0
Bình luận (0)
