Question

si 2senx + cosx = a raiz de 5 calcular x

Answer 1

Respuesta:

Una unción trigonométrica

${sin}^{2} \alpha + {cos}^{2} \alpha = 1$

$cos \alpha = \sqrt{1 - {sin}^{2} \alpha }$

Reemplaza

$2sin \alpha + ( \sqrt{1 - {sin}^{2} \alpha } ) = \sqrt{5}$

$\sqrt{1 - {sin}^{2} \alpha } = \sqrt{5} - 2 sin\alpha$

Elevar al cuadrado

$( { \sqrt{1 - {sin}^{2} \alpha })}^{2} = ( { \sqrt{5} - 2sin \alpha })^{2}$

$1 - {sin}^{2} \alpha = { \sqrt{5} }^{2} - 2( \sqrt{5} )( - 2sin \alpha ) + { ( - 2 \sin \alpha )}^{2}$

$1 - {sin}^{2} \alpha = 5 - 4 \sqrt{5} sin \alpha + 4 {sin}^{2} \alpha$

Terminos iguales lo agrupas

$1 - 5 = {sin}^{2} \alpha - 4 \sqrt{5} sin \alpha + 4 {sin}^{2} \alpha$

$- 4 = 5 {sin}^{2} \alpha - 4 \sqrt{5} \sin \alpha$

Forman un trinomio cuadrado perfecto TCP

$5 {sin}^{2} \alpha - 4 \sqrt{5} sin \alpha + 4 = 0$

$( { \sqrt{5}sin \alpha - 2 })^{2} = 0$

Entonces

$\sqrt{5} sin \alpha = 2$

$sin \alpha = \frac{2}{ \sqrt{5} }$

$\frac{2 \times ( \sqrt{5} )}{ \sqrt{5} \times( \sqrt{5}) } = \frac{2 \sqrt{5} }{5}$

$\alpha = {63.4345}^{0}$