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Popular Trigonometría >

a=((1+sin^2(x)))/((1-sin^2(x)))

Solución

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

Solución

x = arcsin (\frac{- 1 + a}{1 + a}) + 2 πn, x = π + arcsin (- \frac{- 1 + a}{1 + a}) + 2 πn, x = arcsin (- \frac{- 1 + a}{1 + a}) + 2 πn, x = π + arcsin (\frac{- 1 + a}{1 + a}) + 2 πn

Pasos de solución

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

Intercambiar lados

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

Usando el método de sustitución

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

Sea:

sin (x) = u

\frac{1 + u ^{2}}{1 - u ^{2}} = a

\frac{1 + u ^{2}}{1 - u ^{2}} = a : u = \frac{- 1 + a}{1 + a}, u = - \frac{- 1 + a}{1 + a}; a \neq = - 1

\frac{1 + u ^{2}}{1 - u ^{2}} = a

Multiplicar ambos lados por

1 - u^{2}

\frac{1 + u ^{2}}{1 - u ^{2}} = a

Multiplicar ambos lados por

1 - u^{2}

\frac{1 + u ^{2}}{1 - u ^{2}} (1 - u^{2}) = a (1 - u^{2})

Simplificar

1 + u^{2} = a (1 - u^{2})

1 + u^{2} = a (1 - u^{2})

Resolver

1 + u^{2} = a (1 - u^{2}) : u = \frac{- 1 + a}{1 + a}, u = - \frac{- 1 + a}{1 + a}; a \neq = - 1

1 + u^{2} = a (1 - u^{2})

Desplace

1

a la derecha

1 + u^{2} = a (1 - u^{2})

Restar

1

de ambos lados

1 + u^{2} - 1 = a (1 - u^{2}) - 1

Simplificar

u^{2} = a (1 - u^{2}) - 1

u^{2} = a (1 - u^{2}) - 1

Desplace

a (1 - u^{2})

a la izquierda

u^{2} = a (1 - u^{2}) - 1

Restar

a (1 - u^{2})

de ambos lados

u^{2} - a (1 - u^{2}) = a (1 - u^{2}) - 1 - a (1 - u^{2})

Simplificar

u^{2} - a (1 - u^{2}) = - 1

u^{2} - a (1 - u^{2}) = - 1

Expandir

- a (1 - u^{2}) : - a + a u^{2}

- a (1 - u^{2})

Poner los parentesis utilizando:

a (b - c) = ab - a c

a = - a, b = 1, c = u^{2}

= - a \cdot 1 - (- a) u^{2}

Aplicar las reglas de los signos

- (- a) = a

= - 1 \cdot a + a u^{2}

Multiplicar:

1 \cdot a = a

= - a + a u^{2}

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

Desplace

a

a la derecha

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

Sumar

a

a ambos lados

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

Simplificar

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

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

Factorizar

u^{2} + a u^{2} : u^{2} (1 + a)

u^{2} + a u^{2}

Factorizar el termino común

u^{2}

= u^{2} (1 + a)

u^{2} (1 + a) = - 1 + a

Dividir ambos lados entre

1 + a; a \neq = - 1

u^{2} (1 + a) = - 1 + a

Dividir ambos lados entre

1 + a; a \neq = - 1

\frac{u ^{2} ( 1 + a )}{1 + a} = - \frac{1}{1 + a} + \frac{a}{1 + a}; a \neq = - 1

Simplificar

\frac{u ^{2} ( 1 + a )}{1 + a} = - \frac{1}{1 + a} + \frac{a}{1 + a}

Simplificar

\frac{u ^{2} ( 1 + a )}{1 + a} : u^{2}

\frac{u ^{2} ( 1 + a )}{1 + a}

Eliminar los terminos comunes:

1 + a

= u^{2}

Simplificar

- \frac{1}{1 + a} + \frac{a}{1 + a} : \frac{- 1 + a}{1 + a}

- \frac{1}{1 + a} + \frac{a}{1 + a}

Aplicar la regla

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- 1 + a}{1 + a}

u^{2} = \frac{- 1 + a}{1 + a}; a \neq = - 1

u^{2} = \frac{- 1 + a}{1 + a}; a \neq = - 1

u^{2} = \frac{- 1 + a}{1 + a}; a \neq = - 1

Para

x^{2} = f (a)

las soluciones son

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

u = \frac{- 1 + a}{1 + a}, u = - \frac{- 1 + a}{1 + a}; a \neq = - 1

u = \frac{- 1 + a}{1 + a}, u = - \frac{- 1 + a}{1 + a}; a \neq = - 1

Sustituir en la ecuación

u = sin (x)

sin (x) = \frac{- 1 + a}{1 + a}, sin (x) = - \frac{- 1 + a}{1 + a}; a \neq = - 1

sin (x) = \frac{- 1 + a}{1 + a}, sin (x) = - \frac{- 1 + a}{1 + a}; a \neq = - 1

sin (x) = \frac{- 1 + a}{1 + a} : x = arcsin (\frac{- 1 + a}{1 + a}) + 2 πn, x = π + arcsin (- \frac{- 1 + a}{1 + a}) + 2 πn

sin (x) = \frac{- 1 + a}{1 + a}

Aplicar propiedades trigonométricas inversas

sin (x) = \frac{- 1 + a}{1 + a}

Soluciones generales para

sin (x) = \frac{- 1 + a}{1 + a}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (\frac{- 1 + a}{1 + a}) + 2 πn, x = π + arcsin (- \frac{- 1 + a}{1 + a}) + 2 πn

x = arcsin (\frac{- 1 + a}{1 + a}) + 2 πn, x = π + arcsin (- \frac{- 1 + a}{1 + a}) + 2 πn

sin (x) = - \frac{- 1 + a}{1 + a} : x = arcsin (- \frac{- 1 + a}{1 + a}) + 2 πn, x = π + arcsin (\frac{- 1 + a}{1 + a}) + 2 πn

sin (x) = - \frac{- 1 + a}{1 + a}

Aplicar propiedades trigonométricas inversas

sin (x) = - \frac{- 1 + a}{1 + a}

Soluciones generales para

sin (x) = - \frac{- 1 + a}{1 + a}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{- 1 + a}{1 + a}) + 2 πn, x = π + arcsin (\frac{- 1 + a}{1 + a}) + 2 πn

x = arcsin (- \frac{- 1 + a}{1 + a}) + 2 πn, x = π + arcsin (\frac{- 1 + a}{1 + a}) + 2 πn

Combinar toda las soluciones

x = arcsin (\frac{- 1 + a}{1 + a}) + 2 πn, x = π + arcsin (- \frac{- 1 + a}{1 + a}) + 2 πn, x = arcsin (- \frac{- 1 + a}{1 + a}) + 2 πn, x = π + arcsin (\frac{- 1 + a}{1 + a}) + 2 πn

Gráfica

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