[TEX](cos6x+cos2x)+cos4x=0[/TEX]
[TEX]\Leftrightarrow 2cos(\frac{6x+2x}{2}).cos(\frac{6x-2x}{2})+cos4x=0 [/TEX]
[TEX]\Leftrightarrow 2cos4x.cos2x+cos4x=0[/TEX]
[TEX]\Leftrightarrow cos4x(2cos2x+1)=0[/TEX]
[TEX]\Leftrightarrow
\left[\begin{matrix} cos4x=0\\ cos2x=-\frac{1}{2}\end{matrix}\right.
[/TEX]
[TEX]
\Leftrightarrow
\left[\begin{matrix} 4x=\frac{\pi}{2} + k\pi \\ cos2x=cos\frac{2\pi}{3}\end{matrix}\right.
[/TEX]
[TEX]
\Leftrightarrow
\left[\begin{matrix} x=\frac{\pi}{8} + k\frac{\pi}{4}\\ 2x=\frac{2\pi}{3}+k2\pi \\ 2x=-\frac{2\pi}{3}+k2\pi\end{matrix}\right.
[/TEX]
[TEX]
\Leftrightarrow
\left[\begin{matrix} x=\frac{\pi}{8} + k\frac{\pi}{4}\\ x=\frac{\pi}{3}+k\pi \\ x=-\frac{\pi}{3}+k\pi\end{matrix}\right.
[/TEX]
Còn gì thắc mắc về bài này em có thể hỏi bên dưới nha