\(cos5x+cosx-\left(sin6x+sin2x\right)=0\)
\(\Leftrightarrow2cos3x.cos2x-2sin4x.cos2x=0\)
\(\Leftrightarrow2cos2x\left(cos3x-sin4x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos3x=sin4x=cos\left(\dfrac{\pi}{2}-4x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{2}+k\pi\\3x=\dfrac{\pi}{2}-4x+k2\pi\\3x=4x-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Rightarrow...\)
Đúng 1
Bình luận (2)
