Come risolvi $cos 2 x + sin x = 0$ $cos 2x + sin x = 0$ ?

$x = {sin}^{- 1} (- \frac{1}{2}), x = {sin}^{- 1} (1)$ $x=sin^-1(-1/2), x=sin^-1(1)$

Soluzione

$cos 2 x + sin x = 0$ $cos2x+sinx=0$

$cos 2 x = {cos}^{2} x - {sin}^{2} x$ $cos2x=cos^2x-sin^2x$

${cos}^{2} x - {sin}^{2} x + sin x = 0$ $cos^2x-sin^2x+sinx=0$

$1 - {sin}^{2} x - {sin}^{2} x + sin x = 0$ $1-sin^2x-sin^2x+sinx=0$

$1 - 2 {sin}^{2} x + sin x = 0$ $1-2sin^2x+sinx=0$

$- 2 {sin}^{2} x + sin x + 1 = 0$ $-2sin^2x+sinx+1=0$

$2 {sin}^{2} x - sin x - 1 = 0$ $2sin^2x-sinx-1=0$

$2 {sin}^{2} x - 2 sin x + sin x - 1 = 0$ $2sin^2x-2sinx+sinx-1=0$

$2 sin x (sin x - 1) + 1 (sin x - 1)$ $2sinx(sinx-1)+1(sinx-1)$

$(2 sin x + 1) (sin x - 1) = 0$ $(2sinx+1)(sinx-1)=0$

$(2 sin x + 1) = 0, (sin x - 1) = 0$ $(2sinx+1)=0, (sinx-1)=0$

$2 sin x = - 1, sin x = 1$ $2sinx=-1, sinx=1$

$sin x = - \frac{1}{2}, sin x = 1$ $sinx=-1/2, sinx=1$

$x = {sin}^{- 1} (- \frac{1}{2}), x = {sin}^{- 1} (1)$ $x=sin^-1(-1/2), x=sin^-1(1)$

Come risolvi cos2x+sinx=0cos 2x + sin x = 0 ?