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Popular Trigonometría >

tan^2(θ)sec^2(θ)=3

Solución

tan^{2} (θ) sec^{2} (θ) = 3

Solución

θ = 0.85133 \dots + 2 πn, θ = 2 π - 0.85133 \dots + 2 πn, θ = 2.29026 \dots + 2 πn, θ = - 2.29026 \dots + 2 πn

Grados

θ = 48.77764 \dots^{\circ} + 36 0^{\circ} n, θ = 311.22235 \dots^{\circ} + 36 0^{\circ} n, θ = 131.22235 \dots^{\circ} + 36 0^{\circ} n, θ = - 131.22235 \dots^{\circ} + 36 0^{\circ} n

Pasos de solución

tan^{2} (θ) sec^{2} (θ) = 3

Restar

3

de ambos lados

tan^{2} (θ) sec^{2} (θ) - 3 = 0

Re-escribir usando identidades trigonométricas

- 3 + sec^{2} (θ) tan^{2} (θ)

Utilizar la identidad pitagórica:

tan^{2} (x) + 1 = sec^{2} (x)

tan^{2} (x) = sec^{2} (x) - 1

= - 3 + sec^{2} (θ) (sec^{2} (θ) - 1)

- 3 + (- 1 + sec^{2} (θ)) sec^{2} (θ) = 0

Usando el método de sustitución

- 3 + (- 1 + sec^{2} (θ)) sec^{2} (θ) = 0

Sea:

sec (θ) = u

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

- 3 + (- 1 + u^{2}) u^{2} = 0 : u = \frac{1 + 13}{2}, u = - \frac{1 + 13}{2}, u = \frac{1 - 13}{2}, u = - \frac{1 - 13}{2}

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

Desarrollar

- 3 + (- 1 + u^{2}) u^{2} : - 3 - u^{2} + u^{4}

- 3 + (- 1 + u^{2}) u^{2}

= - 3 + u^{2} (- 1 + u^{2})

Expandir

u^{2} (- 1 + u^{2}) : - u^{2} + u^{4}

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

Poner los parentesis utilizando:

a (b + c) = ab + a c

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

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

Aplicar las reglas de los signos

+ (- a) = - a

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

Simplificar

- 1 \cdot u^{2} + u^{2} u^{2} : - u^{2} + u^{4}

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

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

1 \cdot u^{2}

Multiplicar:

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

= u^{2}

u^{2} u^{2} = u^{4}

u^{2} u^{2}

Aplicar las leyes de los exponentes:

a^{b} \cdot a^{c} = a^{b + c}

u^{2} u^{2} = u^{2 + 2}

= u^{2 + 2}

Sumar:

2 + 2 = 4

= u^{4}

= - u^{2} + u^{4}

= - u^{2} + u^{4}

= - 3 - u^{2} + u^{4}

- 3 - u^{2} + u^{4} = 0

Escribir en la forma binómica

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

u^{4} - u^{2} - 3 = 0

Re-escribir la ecuación con

v = u^{2}

v^{2} = u^{4}

v^{2} - v - 3 = 0

Resolver

v^{2} - v - 3 = 0 : v = \frac{1 + 13}{2}, v = \frac{1 - 13}{2}

v^{2} - v - 3 = 0

Resolver con la fórmula general para ecuaciones de segundo grado:

v^{2} - v - 3 = 0

Formula general para ecuaciones de segundo grado:

Para

a = 1, b = - 1, c = - 3

v_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

v_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 3) = 13

(- 1)^{2} - 4 \cdot 1 \cdot (- 3)

Aplicar la regla

- (- a) = a

= (- 1)^{2} + 4 \cdot 1 \cdot 3

(- 1)^{2} = 1

(- 1)^{2}

Aplicar las leyes de los exponentes:

(- a)^{n} = a^{n},

n

es par

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

= 1^{2}

Aplicar la regla

1^{a} = 1

= 1

4 \cdot 1 \cdot 3 = 12

4 \cdot 1 \cdot 3

Multiplicar los numeros:

4 \cdot 1 \cdot 3 = 12

= 12

= 1 + 12

Sumar:

1 + 12 = 13

= 13

v_{1, 2} = \frac{- ( - 1 ) \pm 13}{2 \cdot 1}

Separar las soluciones

v_{1} = \frac{- ( - 1 ) + 13}{2 \cdot 1}, v_{2} = \frac{- ( - 1 ) - 13}{2 \cdot 1}

v = \frac{- ( - 1 ) + 13}{2 \cdot 1} : \frac{1 + 13}{2}

\frac{- ( - 1 ) + 13}{2 \cdot 1}

Aplicar la regla

- (- a) = a

= \frac{1 + 13}{2 \cdot 1}

Multiplicar los numeros:

2 \cdot 1 = 2

= \frac{1 + 13}{2}

v = \frac{- ( - 1 ) - 13}{2 \cdot 1} : \frac{1 - 13}{2}

\frac{- ( - 1 ) - 13}{2 \cdot 1}

Aplicar la regla

- (- a) = a

= \frac{1 - 13}{2 \cdot 1}

Multiplicar los numeros:

2 \cdot 1 = 2

= \frac{1 - 13}{2}

Las soluciones a la ecuación de segundo grado son:

v = \frac{1 + 13}{2}, v = \frac{1 - 13}{2}

v = \frac{1 + 13}{2}, v = \frac{1 - 13}{2}

Sustituir hacia atrás la

v = u^{2},

resolver para

u

Resolver

u^{2} = \frac{1 + 13}{2} : u = \frac{1 + 13}{2}, u = - \frac{1 + 13}{2}

u^{2} = \frac{1 + 13}{2}

Para

x^{2} = f (a)

las soluciones son

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

u = \frac{1 + 13}{2}, u = - \frac{1 + 13}{2}

Resolver

u^{2} = \frac{1 - 13}{2} : u = \frac{1 - 13}{2}, u = - \frac{1 - 13}{2}

u^{2} = \frac{1 - 13}{2}

Para

x^{2} = f (a)

las soluciones son

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

u = \frac{1 - 13}{2}, u = - \frac{1 - 13}{2}

Las soluciones son

u = \frac{1 + 13}{2}, u = - \frac{1 + 13}{2}, u = \frac{1 - 13}{2}, u = - \frac{1 - 13}{2}

Sustituir en la ecuación

u = sec (θ)

sec (θ) = \frac{1 + 13}{2}, sec (θ) = - \frac{1 + 13}{2}, sec (θ) = \frac{1 - 13}{2}, sec (θ) = - \frac{1 - 13}{2}

sec (θ) = \frac{1 + 13}{2}, sec (θ) = - \frac{1 + 13}{2}, sec (θ) = \frac{1 - 13}{2}, sec (θ) = - \frac{1 - 13}{2}

sec (θ) = \frac{1 + 13}{2} : θ = arcsec \frac{1 + 13}{2} + 2 πn, θ = 2 π - arcsec \frac{1 + 13}{2} + 2 πn

sec (θ) = \frac{1 + 13}{2}

Aplicar propiedades trigonométricas inversas

sec (θ) = \frac{1 + 13}{2}

Soluciones generales para

sec (θ) = \frac{1 + 13}{2}

sec (x) = a \Rightarrow x = arcsec (a) + 2 πn, x = 2 π - arcsec (a) + 2 πn

θ = arcsec \frac{1 + 13}{2} + 2 πn, θ = 2 π - arcsec \frac{1 + 13}{2} + 2 πn

θ = arcsec \frac{1 + 13}{2} + 2 πn, θ = 2 π - arcsec \frac{1 + 13}{2} + 2 πn

sec (θ) = - \frac{1 + 13}{2} : θ = arcsec - \frac{1 + 13}{2} + 2 πn, θ = - arcsec - \frac{1 + 13}{2} + 2 πn

sec (θ) = - \frac{1 + 13}{2}

Aplicar propiedades trigonométricas inversas

sec (θ) = - \frac{1 + 13}{2}

Soluciones generales para

sec (θ) = - \frac{1 + 13}{2}

sec (x) = - a \Rightarrow x = arcsec (- a) + 2 πn, x = - arcsec (- a) + 2 πn

θ = arcsec - \frac{1 + 13}{2} + 2 πn, θ = - arcsec - \frac{1 + 13}{2} + 2 πn

θ = arcsec - \frac{1 + 13}{2} + 2 πn, θ = - arcsec - \frac{1 + 13}{2} + 2 πn

sec (θ) = \frac{1 - 13}{2} : θ = arcsec \frac{1 - 13}{2} + 2 πn, θ = - arcsec \frac{1 - 13}{2} + 2 πn

sec (θ) = \frac{1 - 13}{2}

Aplicar propiedades trigonométricas inversas

sec (θ) = \frac{1 - 13}{2}

Soluciones generales para

sec (θ) = \frac{1 - 13}{2}

sec (x) = a \Rightarrow x = arcsec (a) + 2 πn, x = - arcsec (a) + 2 πn

θ = arcsec \frac{1 - 13}{2} + 2 πn, θ = - arcsec \frac{1 - 13}{2} + 2 πn

θ = arcsec \frac{1 - 13}{2} + 2 πn, θ = - arcsec \frac{1 - 13}{2} + 2 πn

sec (θ) = - \frac{1 - 13}{2} : θ = arcsec - \frac{1 - 13}{2} + 2 πn, θ = - arcsec - \frac{1 - 13}{2} + 2 πn

sec (θ) = - \frac{1 - 13}{2}

Aplicar propiedades trigonométricas inversas

sec (θ) = - \frac{1 - 13}{2}

Soluciones generales para

sec (θ) = - \frac{1 - 13}{2}

sec (x) = a \Rightarrow x = arcsec (a) + 2 πn, x = - arcsec (a) + 2 πn

θ = arcsec - \frac{1 - 13}{2} + 2 πn, θ = - arcsec - \frac{1 - 13}{2} + 2 πn

θ = arcsec - \frac{1 - 13}{2} + 2 πn, θ = - arcsec - \frac{1 - 13}{2} + 2 πn

Combinar toda las soluciones

θ = arcsec \frac{1 + 13}{2} + 2 πn, θ = 2 π - arcsec \frac{1 + 13}{2} + 2 πn, θ = arcsec - \frac{1 + 13}{2} + 2 πn, θ = - arcsec - \frac{1 + 13}{2} + 2 πn, θ = arcsec \frac{1 - 13}{2} + 2 πn, θ = - arcsec \frac{1 - 13}{2} + 2 πn, θ = arcsec - \frac{1 - 13}{2} + 2 πn, θ = - arcsec - \frac{1 - 13}{2} + 2 πn

Siendo que la ecuación esta indefinida para:

arcsec \frac{1 - 13}{2} + 2 πn, - arcsec \frac{1 - 13}{2} + 2 πn, arcsec - \frac{1 - 13}{2} + 2 πn, - arcsec - \frac{1 - 13}{2} + 2 πn

θ = arcsec \frac{1 + 13}{2} + 2 πn, θ = 2 π - arcsec \frac{1 + 13}{2} + 2 πn, θ = arcsec - \frac{1 + 13}{2} + 2 πn, θ = - arcsec - \frac{1 + 13}{2} + 2 πn

Mostrar soluciones en forma decimal

θ = 0.85133 \dots + 2 πn, θ = 2 π - 0.85133 \dots + 2 πn, θ = 2.29026 \dots + 2 πn, θ = - 2.29026 \dots + 2 πn

Gráfica

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