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cos^5(x)-sin(x)=0

Solución

cos^{5} (x) - sin (x) = 0

Solución

x = 0.51794 \dots + 2 πn, x = - 2.62364 \dots + 2 πn

Grados

x = 29.67623 \dots^{\circ} + 36 0^{\circ} n, x = - 150.32376 \dots^{\circ} + 36 0^{\circ} n

Pasos de solución

cos^{5} (x) - sin (x) = 0

Sumar

sin (x)

a ambos lados

cos^{5} (x) = sin (x)

Elevar al cuadrado ambos lados

(cos^{5} (x))^{2} = sin^{2} (x)

Restar

sin^{2} (x)

de ambos lados

cos^{10} (x) - sin^{2} (x) = 0

Re-escribir usando identidades trigonométricas

cos^{10} (x) - sin^{2} (x)

Utilizar la identidad pitagórica:

cos^{2} (x) + sin^{2} (x) = 1

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

= cos^{10} (x) - (1 - cos^{2} (x))

- (1 - cos^{2} (x)) : - 1 + cos^{2} (x)

- (1 - cos^{2} (x))

Poner los parentesis

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

Aplicar las reglas de los signos

- (- a) = a, - (a) = - a

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

= cos^{10} (x) - 1 + cos^{2} (x)

- 1 + cos^{10} (x) + cos^{2} (x) = 0

Usando el método de sustitución

- 1 + cos^{10} (x) + cos^{2} (x) = 0

Sea:

cos (x) = u

- 1 + u^{10} + u^{2} = 0

- 1 + u^{10} + u^{2} = 0 : u = 0.75487 \dots, u = - 0.75487 \dots

- 1 + u^{10} + u^{2} = 0

Escribir en la forma binómica

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

u^{10} + u^{2} - 1 = 0

Re-escribir la ecuación con

v = u^{2}

v^{5} = u^{10}

v^{5} + v - 1 = 0

Resolver

v^{5} + v - 1 = 0 : v \approx 0.75487 \dots

v^{5} + v - 1 = 0

Encontrar una solución para

v^{5} + v - 1 = 0

utilizando el método de Newton-Raphson:

v \approx 0.75487 \dots

v^{5} + v - 1 = 0

Definición del método de Newton-Raphson

f (v) = v^{5} + v - 1

Hallar

f^{^{'}} (v) : 5 v^{4} + 1

\frac{d}{d v} (v^{5} + v - 1)

Aplicar la regla de la suma/diferencia:

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

= \frac{d}{d v} (v^{5}) + \frac{d v}{d v} - \frac{d}{d v} (1)

\frac{d}{d v} (v^{5}) = 5 v^{4}

\frac{d}{d v} (v^{5})

Aplicar la regla de la potencia:

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

= 5 v^{5 - 1}

Simplificar

= 5 v^{4}

\frac{d v}{d v} = 1

\frac{d v}{d v}

Aplicar la regla de derivación:

\frac{d v}{d v} = 1

= 1

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

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

Derivada de una constante:

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

= 0

= 5 v^{4} + 1 - 0

Simplificar

= 5 v^{4} + 1

Sea

v_{0} = 1

Calcular

v_{n + 1}

hasta que

Δ v_{n + 1} < 0.000001

v_{1} = 0.83333 \dots : Δ v_{1} = 0.16666 \dots

f (v_{0}) = 1^{5} + 1 - 1 = 1

f^{^{'}} (v_{0}) = 5 \cdot 1^{4} + 1 = 6

v_{1} = 0.83333 \dots

Δ v_{1} = ∣ 0.83333 \dots - 1 ∣ = 0.16666 \dots

Δ v_{1} = 0.16666 \dots

v_{2} = 0.76438 \dots : Δ v_{2} = 0.06895 \dots

f (v_{1}) = 0.83333 \dots^{5} + 0.83333 \dots - 1 = 0.23521 \dots

f^{^{'}} (v_{1}) = 5 \cdot 0.83333 \dots^{4} + 1 = 3.41126 \dots

v_{2} = 0.76438 \dots

Δ v_{2} = ∣ 0.76438 \dots - 0.83333 \dots ∣ = 0.06895 \dots

Δ v_{2} = 0.06895 \dots

v_{3} = 0.75502 \dots : Δ v_{3} = 0.00935 \dots

f (v_{2}) = 0.76438 \dots^{5} + 0.76438 \dots - 1 = 0.02532 \dots

f^{^{'}} (v_{2}) = 5 \cdot 0.76438 \dots^{4} + 1 = 2.70691 \dots

v_{3} = 0.75502 \dots

Δ v_{3} = ∣ 0.75502 \dots - 0.76438 \dots ∣ = 0.00935 \dots

Δ v_{3} = 0.00935 \dots

v_{4} = 0.75487 \dots : Δ v_{4} = 0.00014 \dots

f (v_{3}) = 0.75502 \dots^{5} + 0.75502 \dots - 1 = 0.00038 \dots

f^{^{'}} (v_{3}) = 5 \cdot 0.75502 \dots^{4} + 1 = 2.62485 \dots

v_{4} = 0.75487 \dots

Δ v_{4} = ∣ 0.75487 \dots - 0.75502 \dots ∣ = 0.00014 \dots

Δ v_{4} = 0.00014 \dots

v_{5} = 0.75487 \dots : Δ v_{5} = 3.55234 E - 8

f (v_{4}) = 0.75487 \dots^{5} + 0.75487 \dots - 1 = 9.31989 E - 8

f^{^{'}} (v_{4}) = 5 \cdot 0.75487 \dots^{4} + 1 = 2.62359 \dots

v_{5} = 0.75487 \dots

Δ v_{5} = ∣ 0.75487 \dots - 0.75487 \dots ∣ = 3.55234 E - 8

Δ v_{5} = 3.55234 E - 8

v \approx 0.75487 \dots

Aplicar la división larga

Eq u a t i o n 0 : \frac{v ^{5} + v - 1}{v - 0.75487 \dots} = v^{4} + 0.75487 \dots v^{3} + 0.56984 \dots v^{2} + 0.43015 \dots v + 1.32471 \dots

v^{4} + 0.75487 \dots v^{3} + 0.56984 \dots v^{2} + 0.43015 \dots v + 1.32471 \dots \approx 0

Encontrar una solución para

v^{4} + 0.75487 \dots v^{3} + 0.56984 \dots v^{2} + 0.43015 \dots v + 1.32471 \dots = 0

utilizando el método de Newton-Raphson:

Sin solución para

v \in R

v^{4} + 0.75487 \dots v^{3} + 0.56984 \dots v^{2} + 0.43015 \dots v + 1.32471 \dots = 0

Definición del método de Newton-Raphson

f (v) = v^{4} + 0.75487 \dots v^{3} + 0.56984 \dots v^{2} + 0.43015 \dots v + 1.32471 \dots

Hallar

f^{^{'}} (v) : 4 v^{3} + 2.26463 \dots v^{2} + 1.13968 \dots v + 0.43015 \dots

\frac{d}{d v} (v^{4} + 0.75487 \dots v^{3} + 0.56984 \dots v^{2} + 0.43015 \dots v + 1.32471 \dots)

Aplicar la regla de la suma/diferencia:

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

= \frac{d}{d v} (v^{4}) + \frac{d}{d v} (0.75487 \dots v^{3}) + \frac{d}{d v} (0.56984 \dots v^{2}) + \frac{d}{d v} (0.43015 \dots v) + \frac{d}{d v} (1.32471 \dots)

\frac{d}{d v} (v^{4}) = 4 v^{3}

\frac{d}{d v} (v^{4})

Aplicar la regla de la potencia:

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

= 4 v^{4 - 1}

Simplificar

= 4 v^{3}

\frac{d}{d v} (0.75487 \dots v^{3}) = 2.26463 \dots v^{2}

\frac{d}{d v} (0.75487 \dots v^{3})

Sacar la constante:

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

= 0.75487 \dots \frac{d}{d v} (v^{3})

Aplicar la regla de la potencia:

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

= 0.75487 \dots \cdot 3 v^{3 - 1}

Simplificar

= 2.26463 \dots v^{2}

\frac{d}{d v} (0.56984 \dots v^{2}) = 1.13968 \dots v

\frac{d}{d v} (0.56984 \dots v^{2})

Sacar la constante:

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

= 0.56984 \dots \frac{d}{d v} (v^{2})

Aplicar la regla de la potencia:

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

= 0.56984 \dots \cdot 2 v^{2 - 1}

Simplificar

= 1.13968 \dots v

\frac{d}{d v} (0.43015 \dots v) = 0.43015 \dots

\frac{d}{d v} (0.43015 \dots v)

Sacar la constante:

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

= 0.43015 \dots \frac{d v}{d v}

Aplicar la regla de derivación:

\frac{d v}{d v} = 1

= 0.43015 \dots \cdot 1

Simplificar

= 0.43015 \dots

\frac{d}{d v} (1.32471 \dots) = 0

\frac{d}{d v} (1.32471 \dots)

Derivada de una constante:

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

= 0

= 4 v^{3} + 2.26463 \dots v^{2} + 1.13968 \dots v + 0.43015 \dots + 0

Simplificar

= 4 v^{3} + 2.26463 \dots v^{2} + 1.13968 \dots v + 0.43015 \dots

Sea

v_{0} = - 3

Calcular

v_{n + 1}

hasta que

Δ v_{n + 1} < 0.000001

v_{1} = - 2.27399 \dots : Δ v_{1} = 0.72600 \dots

f (v_{0}) = (- 3)^{4} + 0.75487 \dots (- 3)^{3} + 0.56984 \dots (- 3)^{2} + 0.43015 \dots (- 3) + 1.32471 \dots = 65.78110 \dots

f^{^{'}} (v_{0}) = 4 (- 3)^{3} + 2.26463 \dots (- 3)^{2} + 1.13968 \dots (- 3) + 0.43015 \dots = - 90.60718 \dots

v_{1} = - 2.27399 \dots

Δ v_{1} = ∣ - 2.27399 \dots - (- 3) ∣ = 0.72600 \dots

Δ v_{1} = 0.72600 \dots

v_{2} = - 1.70962 \dots : Δ v_{2} = 0.56437 \dots

f (v_{1}) = (- 2.27399 \dots)^{4} + 0.75487 \dots (- 2.27399 \dots)^{3} + 0.56984 \dots (- 2.27399 \dots)^{2} + 0.43015 \dots (- 2.27399 \dots) + 1.32471 \dots = 21.15650 \dots

f^{^{'}} (v_{1}) = 4 (- 2.27399 \dots)^{3} + 2.26463 \dots (- 2.27399 \dots)^{2} + 1.13968 \dots (- 2.27399 \dots) + 0.43015 \dots = - 37.48682 \dots

v_{2} = - 1.70962 \dots

Δ v_{2} = ∣ - 1.70962 \dots - (- 2.27399 \dots) ∣ = 0.56437 \dots

Δ v_{2} = 0.56437 \dots

v_{3} = - 1.23768 \dots : Δ v_{3} = 0.47193 \dots

f (v_{2}) = (- 1.70962 \dots)^{4} + 0.75487 \dots (- 1.70962 \dots)^{3} + 0.56984 \dots (- 1.70962 \dots)^{2} + 0.43015 \dots (- 1.70962 \dots) + 1.32471 \dots = 7.02564 \dots

f^{^{'}} (v_{2}) = 4 (- 1.70962 \dots)^{3} + 2.26463 \dots (- 1.70962 \dots)^{2} + 1.13968 \dots (- 1.70962 \dots) + 0.43015 \dots = - 14.88684 \dots

v_{3} = - 1.23768 \dots

Δ v_{3} = ∣ - 1.23768 \dots - (- 1.70962 \dots) ∣ = 0.47193 \dots

Δ v_{3} = 0.47193 \dots

v_{4} = - 0.73120 \dots : Δ v_{4} = 0.50648 \dots

f (v_{3}) = (- 1.23768 \dots)^{4} + 0.75487 \dots (- 1.23768 \dots)^{3} + 0.56984 \dots (- 1.23768 \dots)^{2} + 0.43015 \dots (- 1.23768 \dots) + 1.32471 \dots = 2.58063 \dots

f^{^{'}} (v_{3}) = 4 (- 1.23768 \dots)^{3} + 2.26463 \dots (- 1.23768 \dots)^{2} + 1.13968 \dots (- 1.23768 \dots) + 0.43015 \dots = - 5.09520 \dots

v_{4} = - 0.73120 \dots

Δ v_{4} = ∣ - 0.73120 \dots - (- 1.23768 \dots) ∣ = 0.50648 \dots

Δ v_{4} = 0.50648 \dots

v_{5} = 0.99541 \dots : Δ v_{5} = 1.72662 \dots

f (v_{4}) = (- 0.73120 \dots)^{4} + 0.75487 \dots (- 0.73120 \dots)^{3} + 0.56984 \dots (- 0.73120 \dots)^{2} + 0.43015 \dots (- 0.73120 \dots) + 1.32471 \dots = 1.30559 \dots

f^{^{'}} (v_{4}) = 4 (- 0.73120 \dots)^{3} + 2.26463 \dots (- 0.73120 \dots)^{2} + 1.13968 \dots (- 0.73120 \dots) + 0.43015 \dots = - 0.75615 \dots

v_{5} = 0.99541 \dots

Δ v_{5} = ∣ 0.99541 \dots - (- 0.73120 \dots) ∣ = 1.72662 \dots

Δ v_{5} = 1.72662 \dots

v_{6} = 0.47388 \dots : Δ v_{6} = 0.52153 \dots

f (v_{5}) = 0.99541 \dots^{4} + 0.75487 \dots \cdot 0.99541 \dots^{3} + 0.56984 \dots \cdot 0.99541 \dots^{2} + 0.43015 \dots \cdot 0.99541 \dots + 1.32471 \dots = 4.04387 \dots

f^{^{'}} (v_{5}) = 4 \cdot 0.99541 \dots^{3} + 2.26463 \dots \cdot 0.99541 \dots^{2} + 1.13968 \dots \cdot 0.99541 \dots + 0.43015 \dots = 7.75380 \dots

v_{6} = 0.47388 \dots

Δ v_{6} = ∣ 0.47388 \dots - 0.99541 \dots ∣ = 0.52153 \dots

Δ v_{6} = 0.52153 \dots

v_{7} = - 0.46459 \dots : Δ v_{7} = 0.93847 \dots

f (v_{6}) = 0.47388 \dots^{4} + 0.75487 \dots \cdot 0.47388 \dots^{3} + 0.56984 \dots \cdot 0.47388 \dots^{2} + 0.43015 \dots \cdot 0.47388 \dots + 1.32471 \dots = 1.78729 \dots

f^{^{'}} (v_{6}) = 4 \cdot 0.47388 \dots^{3} + 2.26463 \dots \cdot 0.47388 \dots^{2} + 1.13968 \dots \cdot 0.47388 \dots + 0.43015 \dots = 1.90446 \dots

v_{7} = - 0.46459 \dots

Δ v_{7} = ∣ - 0.46459 \dots - 0.47388 \dots ∣ = 0.93847 \dots

Δ v_{7} = 0.93847 \dots

v_{8} = 104.25021 \dots : Δ v_{8} = 104.71480 \dots

f (v_{7}) = (- 0.46459 \dots)^{4} + 0.75487 \dots (- 0.46459 \dots)^{3} + 0.56984 \dots (- 0.46459 \dots)^{2} + 0.43015 \dots (- 0.46459 \dots) + 1.32471 \dots = 1.21875 \dots

f^{^{'}} (v_{7}) = 4 (- 0.46459 \dots)^{3} + 2.26463 \dots (- 0.46459 \dots)^{2} + 1.13968 \dots (- 0.46459 \dots) + 0.43015 \dots = - 0.01163 \dots

v_{8} = 104.25021 \dots

Δ v_{8} = ∣ 104.25021 \dots - (- 0.46459 \dots) ∣ = 104.71480 \dots

Δ v_{8} = 104.71480 \dots

No se puede encontrar solución

La solución es

v \approx 0.75487 \dots

v \approx 0.75487 \dots

Sustituir hacia atrás la

v = u^{2},

resolver para

u

Resolver

u^{2} = 0.75487 \dots : u = 0.75487 \dots, u = - 0.75487 \dots

u^{2} = 0.75487 \dots

Para

x^{2} = f (a)

las soluciones son

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

u = 0.75487 \dots, u = - 0.75487 \dots

Las soluciones son

u = 0.75487 \dots, u = - 0.75487 \dots

Sustituir en la ecuación

u = cos (x)

cos (x) = 0.75487 \dots, cos (x) = - 0.75487 \dots

cos (x) = 0.75487 \dots, cos (x) = - 0.75487 \dots

cos (x) = 0.75487 \dots : x = arccos (0.75487 \dots) + 2 πn, x = 2 π - arccos (0.75487 \dots) + 2 πn

cos (x) = 0.75487 \dots

Aplicar propiedades trigonométricas inversas

cos (x) = 0.75487 \dots

Soluciones generales para

cos (x) = 0.75487 \dots

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (0.75487 \dots) + 2 πn, x = 2 π - arccos (0.75487 \dots) + 2 πn

x = arccos (0.75487 \dots) + 2 πn, x = 2 π - arccos (0.75487 \dots) + 2 πn

cos (x) = - 0.75487 \dots : x = arccos (- 0.75487 \dots) + 2 πn, x = - arccos (- 0.75487 \dots) + 2 πn

cos (x) = - 0.75487 \dots

Aplicar propiedades trigonométricas inversas

cos (x) = - 0.75487 \dots

Soluciones generales para

cos (x) = - 0.75487 \dots

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (- 0.75487 \dots) + 2 πn, x = - arccos (- 0.75487 \dots) + 2 πn

x = arccos (- 0.75487 \dots) + 2 πn, x = - arccos (- 0.75487 \dots) + 2 πn

Combinar toda las soluciones

x = arccos (0.75487 \dots) + 2 πn, x = 2 π - arccos (0.75487 \dots) + 2 πn, x = arccos (- 0.75487 \dots) + 2 πn, x = - arccos (- 0.75487 \dots) + 2 πn

Verificar las soluciones sustituyendo en la ecuación original

Verificar las soluciones sustituyéndolas en

cos^{5} (x) - sin (x) = 0

Quitar las que no concuerden con la ecuación.

Verificar la solución

arccos (0.75487 \dots) + 2 πn :

Verdadero

arccos (0.75487 \dots) + 2 πn

Sustituir

n = 1

arccos (0.75487 \dots) + 2 π 1

Multiplicar

cos^{5} (x) - sin (x) = 0

por

x = arccos (0.75487 \dots) + 2 π 1

cos^{5} (arccos (0.75487 \dots) + 2 π 1) - sin (arccos (0.75487 \dots) + 2 π 1) = 0

Simplificar

0 = 0

\Rightarrow Verdadero

Verificar la solución

2 π - arccos (0.75487 \dots) + 2 πn :

Falso

2 π - arccos (0.75487 \dots) + 2 πn

Sustituir

n = 1

2 π - arccos (0.75487 \dots) + 2 π 1

Multiplicar

cos^{5} (x) - sin (x) = 0

por

x = 2 π - arccos (0.75487 \dots) + 2 π 1

cos^{5} (2 π - arccos (0.75487 \dots) + 2 π 1) - sin (2 π - arccos (0.75487 \dots) + 2 π 1) = 0

Simplificar

0.99019 \dots = 0

\Rightarrow Falso

Verificar la solución

arccos (- 0.75487 \dots) + 2 πn :

Falso

arccos (- 0.75487 \dots) + 2 πn

Sustituir

n = 1

arccos (- 0.75487 \dots) + 2 π 1

Multiplicar

cos^{5} (x) - sin (x) = 0

por

x = arccos (- 0.75487 \dots) + 2 π 1

cos^{5} (arccos (- 0.75487 \dots) + 2 π 1) - sin (arccos (- 0.75487 \dots) + 2 π 1) = 0

Simplificar

- 0.99019 \dots = 0

\Rightarrow Falso

Verificar la solución

- arccos (- 0.75487 \dots) + 2 πn :

Verdadero

- arccos (- 0.75487 \dots) + 2 πn

Sustituir

n = 1

- arccos (- 0.75487 \dots) + 2 π 1

Multiplicar

cos^{5} (x) - sin (x) = 0

por

x = - arccos (- 0.75487 \dots) + 2 π 1

cos^{5} (- arccos (- 0.75487 \dots) + 2 π 1) - sin (- arccos (- 0.75487 \dots) + 2 π 1) = 0

Simplificar

0 = 0

\Rightarrow Verdadero

x = arccos (0.75487 \dots) + 2 πn, x = - arccos (- 0.75487 \dots) + 2 πn

Mostrar soluciones en forma decimal

x = 0.51794 \dots + 2 πn, x = - 2.62364 \dots + 2 πn

Ejemplos populares

solvefor x,y= 1/(5sin(2x))resolver para

x, y = \frac{1}{5 sin ( 2 x )}

cos(θ)= 3/(sqrt(35))

cos (θ) = \frac{3}{35}

csc^2(θ)-3csc(θ)+2=0

csc^{2} (θ) - 3 csc (θ) + 2 = 0

sin(3x)=sqrt(2)

sin (3 x) = 2

4*cos(x)+5=0

4 \cdot cos (x) + 5 = 0