4cosX * cos2X= cos2X/sinX

$4cosx*cos2x=\frac{cos2x}{sinx}$
$4cosx*cos2x-cos2x*\frac{1}{sinx}=0$
$cos2x(4cosx-\frac{1}{sinx})=0$
1 ур-ние
$cos2x=0; sinx\ne 0$
$2x=\frac{\pi}{2}+\pi n, sinx\ne 0, n\in Z$
$x=\frac{\pi}{4}+\frac{\pi n}{2}, n\in Z$
2 ур-ние
$4cosx=\frac{1}{sinx}$
$4cosxsinx=1$
$cosxsinx=\frac{1}{4}$
$\frac{1}{2}(sin2x)=\frac{1}{4}$
$sin2x=\frac{1}{2}$
$2x=(-1)^n arcsin(\frac{1}{2})+\pi n, n\in Z$
$x=\frac{1}{2}((-1)^n \frac{\pi}{6}+\pi n), n\in Z$
Ответ: $x_1=\frac{\pi}{4}+\frac{\pi n}{2}, n\in Z$
$x_2=\frac{1}{2}((-1)^n \frac{\pi}{6}+\pi n), n\in Z$