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(sec^2(x)+1)(sec^2(x)-1)=tan(x)

Solución

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

Solución

x = πn, x = 0.42567 \dots + πn

Grados

x = 0^{\circ} + 18 0^{\circ} n, x = 24.38942 \dots^{\circ} + 18 0^{\circ} n

Pasos de solución

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

Restar

tan (x)

de ambos lados

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

Re-escribir usando identidades trigonométricas

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

Utilizar la identidad pitagórica:

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

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

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

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

Factorizar

- tan (x) + (1 + sec^{2} (x)) tan^{2} (x) : tan (x) (- 1 + tan (x) (sec^{2} (x) + 1))

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

Aplicar las leyes de los exponentes:

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

tan^{2} (x) = tan (x) tan (x)

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

Factorizar el termino común

tan (x)

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

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

Resolver cada parte por separado

tan (x) = 0 or - 1 + tan (x) (sec^{2} (x) + 1) = 0

tan (x) = 0 : x = πn

tan (x) = 0

Soluciones generales para

tan (x) = 0

tan (x)

tabla de valores periódicos con

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = 0 + πn

x = 0 + πn

Resolver

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = πn

- 1 + tan (x) (sec^{2} (x) + 1) = 0 : x = arctan (0.45339 \dots) + πn

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

Re-escribir usando identidades trigonométricas

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

Utilizar la identidad pitagórica:

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

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

Simplificar

1 + tan^{2} (x) + 1 : tan^{2} (x) + 2

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

Agrupar términos semejantes

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

Sumar:

1 + 1 = 2

= tan^{2} (x) + 2

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

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

Usando el método de sustitución

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

Sea:

tan (x) = u

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

- 1 + (2 + u^{2}) u = 0 : u \approx 0.45339 \dots

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

Desarrollar

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

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

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

Expandir

u (2 + u^{2}) : 2 u + u^{3}

u (2 + u^{2})

Poner los parentesis utilizando:

a (b + c) = ab + a c

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

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

= 2 u + u^{2} u

u^{2} u = u^{3}

u^{2} u

Aplicar las leyes de los exponentes:

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

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

= u^{2 + 1}

Sumar:

2 + 1 = 3

= u^{3}

= 2 u + u^{3}

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

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

Escribir en la forma binómica

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

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

Encontrar una solución para

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

utilizando el método de Newton-Raphson:

u \approx 0.45339 \dots

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

Definición del método de Newton-Raphson

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

Hallar

f^{^{'}} (u) : 3 u^{2} + 2

\frac{d}{d u} (u^{3} + 2 u - 1)

Aplicar la regla de la suma/diferencia:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d}{d u} (u^{3}) = 3 u^{2}

\frac{d}{d u} (u^{3})

Aplicar la regla de la potencia:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 3 u^{3 - 1}

Simplificar

= 3 u^{2}

\frac{d}{d u} (2 u) = 2

\frac{d}{d u} (2 u)

Sacar la constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d u}{d u}

Aplicar la regla de derivación:

\frac{d u}{d u} = 1

= 2 \cdot 1

Simplificar

= 2

\frac{d}{d u} (1) = 0

\frac{d}{d u} (1)

Derivada de una constante:

\frac{d}{d x} (a) = 0

= 0

= 3 u^{2} + 2 - 0

Simplificar

= 3 u^{2} + 2

Sea

u_{0} = 1

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = 0.6 : Δ u_{1} = 0.4

f (u_{0}) = 1^{3} + 2 \cdot 1 - 1 = 2

f^{^{'}} (u_{0}) = 3 \cdot 1^{2} + 2 = 5

u_{1} = 0.6

Δ u_{1} = ∣ 0.6 - 1 ∣ = 0.4

Δ u_{1} = 0.4

u_{2} = 0.46493 \dots : Δ u_{2} = 0.13506 \dots

f (u_{1}) = 0. 6^{3} + 2 \cdot 0.6 - 1 = 0.416

f^{^{'}} (u_{1}) = 3 \cdot 0. 6^{2} + 2 = 3.08

u_{2} = 0.46493 \dots

Δ u_{2} = ∣ 0.46493 \dots - 0.6 ∣ = 0.13506 \dots

Δ u_{2} = 0.13506 \dots

u_{3} = 0.45346 \dots : Δ u_{3} = 0.01146 \dots

f (u_{2}) = 0.46493 \dots^{3} + 2 \cdot 0.46493 \dots - 1 = 0.03037 \dots

f^{^{'}} (u_{2}) = 3 \cdot 0.46493 \dots^{2} + 2 = 2.64849 \dots

u_{3} = 0.45346 \dots

Δ u_{3} = ∣ 0.45346 \dots - 0.46493 \dots ∣ = 0.01146 \dots

Δ u_{3} = 0.01146 \dots

u_{4} = 0.45339 \dots : Δ u_{4} = 0.00006 \dots

f (u_{3}) = 0.45346 \dots^{3} + 2 \cdot 0.45346 \dots - 1 = 0.00018 \dots

f^{^{'}} (u_{3}) = 3 \cdot 0.45346 \dots^{2} + 2 = 2.61689 \dots

u_{4} = 0.45339 \dots

Δ u_{4} = ∣ 0.45339 \dots - 0.45346 \dots ∣ = 0.00006 \dots

Δ u_{4} = 0.00006 \dots

u_{5} = 0.45339 \dots : Δ u_{5} = 2.5125 E - 9

f (u_{4}) = 0.45339 \dots^{3} + 2 \cdot 0.45339 \dots - 1 = 6.57449 E - 9

f^{^{'}} (u_{4}) = 3 \cdot 0.45339 \dots^{2} + 2 = 2.61670 \dots

u_{5} = 0.45339 \dots

Δ u_{5} = ∣ 0.45339 \dots - 0.45339 \dots ∣ = 2.5125 E - 9

Δ u_{5} = 2.5125 E - 9

u \approx 0.45339 \dots

Aplicar la división larga

Eq u a t i o n 0 : \frac{u ^{3} + 2 u - 1}{u - 0.45339 \dots} = u^{2} + 0.45339 \dots u + 2.20556 \dots

u^{2} + 0.45339 \dots u + 2.20556 \dots \approx 0

Encontrar una solución para

u^{2} + 0.45339 \dots u + 2.20556 \dots = 0

utilizando el método de Newton-Raphson:

Sin solución para

u \in R

u^{2} + 0.45339 \dots u + 2.20556 \dots = 0

Definición del método de Newton-Raphson

f (u) = u^{2} + 0.45339 \dots u + 2.20556 \dots

Hallar

f^{^{'}} (u) : 2 u + 0.45339 \dots

\frac{d}{d u} (u^{2} + 0.45339 \dots u + 2.20556 \dots)

Aplicar la regla de la suma/diferencia:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (u^{2}) + \frac{d}{d u} (0.45339 \dots u) + \frac{d}{d u} (2.20556 \dots)

\frac{d}{d u} (u^{2}) = 2 u

\frac{d}{d u} (u^{2})

Aplicar la regla de la potencia:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 u^{2 - 1}

Simplificar

= 2 u

\frac{d}{d u} (0.45339 \dots u) = 0.45339 \dots

\frac{d}{d u} (0.45339 \dots u)

Sacar la constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 0.45339 \dots \frac{d u}{d u}

Aplicar la regla de derivación:

\frac{d u}{d u} = 1

= 0.45339 \dots \cdot 1

Simplificar

= 0.45339 \dots

\frac{d}{d u} (2.20556 \dots) = 0

\frac{d}{d u} (2.20556 \dots)

Derivada de una constante:

\frac{d}{d x} (a) = 0

= 0

= 2 u + 0.45339 \dots + 0

Simplificar

= 2 u + 0.45339 \dots

Sea

u_{0} = - 5

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = - 2.38770 \dots : Δ u_{1} = 2.61229 \dots

f (u_{0}) = (- 5)^{2} + 0.45339 \dots (- 5) + 2.20556 \dots = 24.93858 \dots

f^{^{'}} (u_{0}) = 2 (- 5) + 0.45339 \dots = - 9.54660 \dots

u_{1} = - 2.38770 \dots

Δ u_{1} = ∣ - 2.38770 \dots - (- 5) ∣ = 2.61229 \dots

Δ u_{1} = 2.61229 \dots

u_{2} = - 0.80877 \dots : Δ u_{2} = 1.57892 \dots

f (u_{1}) = (- 2.38770 \dots)^{2} + 0.45339 \dots (- 2.38770 \dots) + 2.20556 \dots = 6.82410 \dots

f^{^{'}} (u_{1}) = 2 (- 2.38770 \dots) + 0.45339 \dots = - 4.32200 \dots

u_{2} = - 0.80877 \dots

Δ u_{2} = ∣ - 0.80877 \dots - (- 2.38770 \dots) ∣ = 1.57892 \dots

Δ u_{2} = 1.57892 \dots

u_{3} = 1.33267 \dots : Δ u_{3} = 2.14145 \dots

f (u_{2}) = (- 0.80877 \dots)^{2} + 0.45339 \dots (- 0.80877 \dots) + 2.20556 \dots = 2.49299 \dots

f^{^{'}} (u_{2}) = 2 (- 0.80877 \dots) + 0.45339 \dots = - 1.16416 \dots

u_{3} = 1.33267 \dots

Δ u_{3} = ∣ 1.33267 \dots - (- 0.80877 \dots) ∣ = 2.14145 \dots

Δ u_{3} = 2.14145 \dots

u_{4} = - 0.13773 \dots : Δ u_{4} = 1.47040 \dots

f (u_{3}) = 1.33267 \dots^{2} + 0.45339 \dots \cdot 1.33267 \dots + 2.20556 \dots = 4.58582 \dots

f^{^{'}} (u_{3}) = 2 \cdot 1.33267 \dots + 0.45339 \dots = 3.11874 \dots

u_{4} = - 0.13773 \dots

Δ u_{4} = ∣ - 0.13773 \dots - 1.33267 \dots ∣ = 1.47040 \dots

Δ u_{4} = 1.47040 \dots

u_{5} = - 12.28878 \dots : Δ u_{5} = 12.15105 \dots

f (u_{4}) = (- 0.13773 \dots)^{2} + 0.45339 \dots (- 0.13773 \dots) + 2.20556 \dots = 2.16209 \dots

f^{^{'}} (u_{4}) = 2 (- 0.13773 \dots) + 0.45339 \dots = 0.17793 \dots

u_{5} = - 12.28878 \dots

Δ u_{5} = ∣ - 12.28878 \dots - (- 0.13773 \dots) ∣ = 12.15105 \dots

Δ u_{5} = 12.15105 \dots

u_{6} = - 6.16844 \dots : Δ u_{6} = 6.12033 \dots

f (u_{5}) = (- 12.28878 \dots)^{2} + 0.45339 \dots (- 12.28878 \dots) + 2.20556 \dots = 147.64808 \dots

f^{^{'}} (u_{5}) = 2 (- 12.28878 \dots) + 0.45339 \dots = - 24.12417 \dots

u_{6} = - 6.16844 \dots

Δ u_{6} = ∣ - 6.16844 \dots - (- 12.28878 \dots) ∣ = 6.12033 \dots

Δ u_{6} = 6.12033 \dots

u_{7} = - 3.01629 \dots : Δ u_{7} = 3.15214 \dots

f (u_{6}) = (- 6.16844 \dots)^{2} + 0.45339 \dots (- 6.16844 \dots) + 2.20556 \dots = 37.45853 \dots

f^{^{'}} (u_{6}) = 2 (- 6.16844 \dots) + 0.45339 \dots = - 11.88349 \dots

u_{7} = - 3.01629 \dots

Δ u_{7} = ∣ - 3.01629 \dots - (- 6.16844 \dots) ∣ = 3.15214 \dots

Δ u_{7} = 3.15214 \dots

u_{8} = - 1.23538 \dots : Δ u_{8} = 1.78090 \dots

f (u_{7}) = (- 3.01629 \dots)^{2} + 0.45339 \dots (- 3.01629 \dots) + 2.20556 \dots = 9.93603 \dots

f^{^{'}} (u_{7}) = 2 (- 3.01629 \dots) + 0.45339 \dots = - 5.57919 \dots

u_{8} = - 1.23538 \dots

Δ u_{8} = ∣ - 1.23538 \dots - (- 3.01629 \dots) ∣ = 1.78090 \dots

Δ u_{8} = 1.78090 \dots

u_{9} = 0.33676 \dots : Δ u_{9} = 1.57215 \dots

f (u_{8}) = (- 1.23538 \dots)^{2} + 0.45339 \dots (- 1.23538 \dots) + 2.20556 \dots = 3.17163 \dots

f^{^{'}} (u_{8}) = 2 (- 1.23538 \dots) + 0.45339 \dots = - 2.01738 \dots

u_{9} = 0.33676 \dots

Δ u_{9} = ∣ 0.33676 \dots - (- 1.23538 \dots) ∣ = 1.57215 \dots

Δ u_{9} = 1.57215 \dots

u_{10} = - 1.85651 \dots : Δ u_{10} = 2.19328 \dots

f (u_{9}) = 0.33676 \dots^{2} + 0.45339 \dots \cdot 0.33676 \dots + 2.20556 \dots = 2.47166 \dots

f^{^{'}} (u_{9}) = 2 \cdot 0.33676 \dots + 0.45339 \dots = 1.12692 \dots

u_{10} = - 1.85651 \dots

Δ u_{10} = ∣ - 1.85651 \dots - 0.33676 \dots ∣ = 2.19328 \dots

Δ u_{10} = 2.19328 \dots

u_{11} = - 0.38074 \dots : Δ u_{11} = 1.47577 \dots

f (u_{10}) = (- 1.85651 \dots)^{2} + 0.45339 \dots (- 1.85651 \dots) + 2.20556 \dots = 4.81048 \dots

f^{^{'}} (u_{10}) = 2 (- 1.85651 \dots) + 0.45339 \dots = - 3.25963 \dots

u_{11} = - 0.38074 \dots

Δ u_{11} = ∣ - 0.38074 \dots - (- 1.85651 \dots) ∣ = 1.47577 \dots

Δ u_{11} = 1.47577 \dots

u_{12} = 6.68826 \dots : Δ u_{12} = 7.06901 \dots

f (u_{11}) = (- 0.38074 \dots)^{2} + 0.45339 \dots (- 0.38074 \dots) + 2.20556 \dots = 2.17790 \dots

f^{^{'}} (u_{11}) = 2 (- 0.38074 \dots) + 0.45339 \dots = - 0.30809 \dots

u_{12} = 6.68826 \dots

Δ u_{12} = ∣ 6.68826 \dots - (- 0.38074 \dots) ∣ = 7.06901 \dots

Δ u_{12} = 7.06901 \dots

No se puede encontrar solución

La solución es

u \approx 0.45339 \dots

Sustituir en la ecuación

u = tan (x)

tan (x) \approx 0.45339 \dots

tan (x) \approx 0.45339 \dots

tan (x) = 0.45339 \dots : x = arctan (0.45339 \dots) + πn

tan (x) = 0.45339 \dots

Aplicar propiedades trigonométricas inversas

tan (x) = 0.45339 \dots

Soluciones generales para

tan (x) = 0.45339 \dots

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

x = arctan (0.45339 \dots) + πn

x = arctan (0.45339 \dots) + πn

Combinar toda las soluciones

x = arctan (0.45339 \dots) + πn

Combinar toda las soluciones

x = πn, x = arctan (0.45339 \dots) + πn

Mostrar soluciones en forma decimal

x = πn, x = 0.42567 \dots + πn

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\frac{cot ( x )}{csc ( + 1 )} = \frac{csc ( - 1 )}{cot ( x )}

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tan(b)=3

tan (b) = 3