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(2sin^2(x)-1)(3tan^2(x)-1)=0

Solución

(2 sin^{2} (x) - 1) (3 tan^{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, x = 0.52359 \dots + πn, x = - 0.52359 \dots + πn

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, x = 3 0^{\circ} + 18 0^{\circ} n, x = - 3 0^{\circ} + 18 0^{\circ} n

Pasos de solución

(2 sin^{2} (x) - 1) (3 tan^{2} (x) - 1) = 0

Resolver cada parte por separado

2 sin^{2} (x) - 1 = 0 or 3 tan^{2} (x) - 1 = 0

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

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

Usando el método de sustitución

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

Sea:

sin (x) = u

2 u^{2} - 1 = 0

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

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

3 tan^{2} (x) - 1 = 0 : x = arctan (\frac{1}{3}) + πn, x = arctan (- \frac{1}{3}) + πn

3 tan^{2} (x) - 1 = 0

Usando el método de sustitución

3 tan^{2} (x) - 1 = 0

Sea:

tan (x) = u

3 u^{2} - 1 = 0

3 u^{2} - 1 = 0 : u = \frac{1}{3}, u = - \frac{1}{3}

3 u^{2} - 1 = 0

Mover

1

al lado derecho

3 u^{2} - 1 = 0

Sumar

1

a ambos lados

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

Simplificar

3 u^{2} = 1

3 u^{2} = 1

Dividir ambos lados entre

3

3 u^{2} = 1

Dividir ambos lados entre

3

\frac{3 u ^{2}}{3} = \frac{1}{3}

Simplificar

u^{2} = \frac{1}{3}

u^{2} = \frac{1}{3}

Para

x^{2} = f (a)

las soluciones son

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

u = \frac{1}{3}, u = - \frac{1}{3}

Sustituir en la ecuación

u = tan (x)

tan (x) = \frac{1}{3}, tan (x) = - \frac{1}{3}

tan (x) = \frac{1}{3}, tan (x) = - \frac{1}{3}

tan (x) = \frac{1}{3} : x = arctan (\frac{1}{3}) + πn

tan (x) = \frac{1}{3}

Aplicar propiedades trigonométricas inversas

tan (x) = \frac{1}{3}

Soluciones generales para

tan (x) = \frac{1}{3}

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan (\frac{1}{3}) + πn

x = arctan (\frac{1}{3}) + πn

tan (x) = - \frac{1}{3} : x = arctan (- \frac{1}{3}) + πn

tan (x) = - \frac{1}{3}

Aplicar propiedades trigonométricas inversas

tan (x) = - \frac{1}{3}

Soluciones generales para

tan (x) = - \frac{1}{3}

tan (x) = - a \Rightarrow x = arctan (- a) + πn

x = arctan (- \frac{1}{3}) + πn

x = arctan (- \frac{1}{3}) + πn

Combinar toda las soluciones

x = arctan (\frac{1}{3}) + πn, x = arctan (- \frac{1}{3}) + 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, x = arctan (\frac{1}{3}) + πn, x = arctan (- \frac{1}{3}) + πn

Mostrar soluciones en forma decimal

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, x = 0.52359 \dots + πn, x = - 0.52359 \dots + πn

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cos (x) + 2 sin (x) = 2

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3cos(2t)+4=1

3 cos (2 t) + 4 = 1

sin^2(x)=4-2cos^2(x)

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