Question

\(3\cos2x+4\cos^3x-\cos3x=0\)

Answer 1

\(3cos2x+4cos^3x-cos3x=0\\ \Leftrightarrow3cos2x+4cos^3x-\left(4cos^3x-3cosx\right)=0\\ \Leftrightarrow cos2x=cos\left(\pi-x\right)\\ \Leftrightarrow\left[{}\begin{matrix}2x=\pi-x+k2\pi\\2x=x-\pi+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+\frac{k2\pi}{3}\\x=-\pi+k2\pi\end{matrix}\right.\\ \Leftrightarrow x=\frac{\pi}{3}+\frac{k2\pi}{3}\)

Answer 2

bạn ơi, sao cos3x chuyển thành 4cos³x-3cosx vậy ạ