Februar 22, 2020

von Madelina

Wie löst man ${cos}^{2} x - cos x = 0$ $cos ^ 2x - cosx = 0$ ?

Antworten:

$x = 0^{o}$ $x=0^o$ ; $x = 90^{o}$ $x=90^o$

Erläuterung:

${cos}^{2} x - cos x = 0$ $Cos^2 x-Cos x=0$
$cos x (1 - cos x) = 0$ $Cos x(1-Cos x)=0$
$cos x = 0; cos x = 1$ $Cosx=0 ; Cos x=1$
$x = 0^{o}$ $x=0^o$ ; $x = 90^{o}$ $x=90^o$

Test:
(ein)
$x = 0^{o}$ $x=0^o$
${cos}^{2} 0$ $Cos^2 0$ - $cos 0$ $Cos 0$
$1^{2} - 1 = 0$ $1^2 -1 = 0$
(B)
$x = 0^{o}$ $x=0^o$
${cos}^{2} 90$ $Cos^2 90$ - $cos 90$ $Cos 90$
$0^{2} - 0 = 0$ $0^2 -0= 0$