Nullpunkt for funksjonen: 5*e^(-x/3)*sin(2*x) = 0 --> sin(2*x)=0 ---> x=(pi/2) * n
Kan noen forklare meg hvordan sin(2*x)=0 medfører at x=(pi/2) * n ?
-på forhånd takk!
trigonometriske likninger og funksjoner
Moderators: Aleks855, Gustav, Nebuchadnezzar, Janhaa, DennisChristensen, Emilga
$\sin(x)$ har nullpunkter for $x \in \{0, \pi, 2\pi, 3\pi, \ldots\}$.
$\sin(2x)$ har da nullpuntker for $2x \in \{0, \pi, 2\pi, 3\pi, \ldots\}$.
$\sin(2x)$ har da nullpuntker for $2x \in \{0, \pi, 2\pi, 3\pi, \ldots\}$.
-
Dolandyret
- Lagrange
- Posts: 1264
- Joined: 04/10-2015 22:21
Aleks855 wrote:$\sin(x)$ har nullpunkter for $x \in \{0, \pi, 2\pi, 3\pi, \ldots\}$.
$\sin(2x)$ har da nullpuntker for $2x \in \{0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi, \ldots\}$.
"I want to die peacefully in my sleep like my grandfather, not screaming in terror like his passengers."