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9(2sin^2(x)-1)=0

Solución

9 (2 sin^{2} (x) - 1) = 0

Solución

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

+1

Grados

x = 4 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n, x = 22 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n

Pasos de solución

9 (2 sin^{2} (x) - 1) = 0

Usando el método de sustitución

9 (2 sin^{2} (x) - 1) = 0

Sea:

sin (x) = u

9 (2 u^{2} - 1) = 0

9 (2 u^{2} - 1) = 0 : u = \frac{1}{2}, u = - \frac{1}{2}

9 (2 u^{2} - 1) = 0

Dividir ambos lados entre

9

9 (2 u^{2} - 1) = 0

Dividir ambos lados entre

9

\frac{9 ( 2 u ^{2} - 1 )}{9} = \frac{0}{9}

Simplificar

2 u^{2} - 1 = 0

2 u^{2} - 1 = 0

Mover

1

al lado derecho

2 u^{2} - 1 = 0

Sumar

1

a ambos lados

2 u^{2} - 1 + 1 = 0 + 1

Simplificar

2 u^{2} = 1

2 u^{2} = 1

Dividir ambos lados entre

2

2 u^{2} = 1

Dividir ambos lados entre

2

\frac{2 u ^{2}}{2} = \frac{1}{2}

Simplificar

u^{2} = \frac{1}{2}

u^{2} = \frac{1}{2}

Para

x^{2} = f (a)

las soluciones son

x = f (a), - f (a)

u = \frac{1}{2}, u = - \frac{1}{2}

Sustituir en la ecuación

u = sin (x)

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

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

sin (x) = \frac{1}{2} : x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

sin (x) = \frac{1}{2}

Soluciones generales para

sin (x) = \frac{1}{2}

tabla de valores periódicos con

2 πn

intervalos:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

sin (x) = - \frac{1}{2} : x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

sin (x) = - \frac{1}{2}

Soluciones generales para

sin (x) = - \frac{1}{2}

tabla de valores periódicos con

2 πn

intervalos:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Combinar toda las soluciones

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

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sin(9x)-cos(x)=0

sin (9 x) - cos (x) = 0

cos(x)+2cos^2(x)=1

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- \frac{2}{3} = cos (x)