Wie löst man $cos x - cos 2 x = 0$ $cos x - cos 2x = 0$ ?

Verwenden Sie die Eigenschaft: $cos 2 A = 2 {cos}^{- 2} A - 1$ $cos 2A=2cos^-2A-1$

$cos x - (2 {cos}^{2} x - 1) = 0$ $cosx-(2cos^2x -1)=0$

$- 1 [cos x - 2 {cos}^{2} x + 1] = 0$ $-1 [cosx -2cos^2x+1]=0$

$2 {cos}^{2} x - cos x - 1 = 0$ $2cos^2x-cosx-1=0$

$(2 cos x + 1) (cos x - 1) = 0$ $(2cosx+1)(cosx-1)=0$

$cos x = - \frac{1}{2} or cos x = 1$ $cosx=-1/2 or cos x=1$

$x = {cos}^{- 1} (- \frac{1}{2}) or x = {cos}^{- 1} 1$ $x=cos^-1(-1/2) or x=cos^-1 1$

$x = \pm \frac{2 π}{3} + 2 π n or x = 0 + 2 π n$ $x=+- (2pi)/3 + 2pin or x=0+2pin$

$S = {\pm \frac{2 π}{3} + 2 π n, 0 + 2 π n}$ $S={+- (2pi)/3 + 2pin , 0+2pin}$

Wie löst man cos x - cos 2x = 0 cosx−cos2x=0cos x - cos 2x = 0 ?