$pt \Leftrightarrow cos6x+cos4x=\frac{\sqrt3}{2}.cos2x+\frac{1}{2}sin2 x+\frac{\sqrt3}{2} $$\Leftrightarrow 2.cos5x.cosx=cos(2x-\frac{\pi}{6})+cos\frac{\pi}{6}$$\Leftrightarrow 2.cos5x.cosx=2.cosx.cos(x-\frac{\pi}{6})$$+) cosx=0 \Leftrightarrow x=\frac{\pi}{2}+k2\pi$$+) cos5x=cos(x-\frac{\pi}2)$Đến đây có thể tự giải rồi
$pt \Leftrightarrow
\cos6x+
\cos4x=\
dfrac{\sqrt3}{2}
\cos2x+\
dfrac{1}{2}
\sin2 x+\frac{\sqrt3}{2}
\\\Leftrightarrow 2
\cos5x
\cos
x=
\cos
\left(2x-\
dfrac{\pi}{6}
\right)+
\cos\
dfrac{\pi}{6}
\\\Leftrightarrow 2
\cos5x
\cos
x=2
\cos
x
\cos
\left(x-\
dfrac{\pi}{6}
\right)
\\\Leftrightarrow \
lef
t[\begin{arra
y}{
1}\
cos x=0 \\\cos5x=
\cos
\left(x-\
dfrac{\pi}
{2
}\right)
\end{array}\right.$Đến đây có thể tự giải rồi
.