English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5sec^2(θ)sin(θ)-cos(θ)=0

Solución

5 sec^{2} (θ) sin (θ) - cos (θ) = 0

Solución

θ = 0.19049 \dots + 2 πn, θ = - 2.95110 \dots + 2 πn

Grados

θ = 10.91437 \dots^{\circ} + 36 0^{\circ} n, θ = - 169.08562 \dots^{\circ} + 36 0^{\circ} n

Pasos de solución

5 sec^{2} (θ) sin (θ) - cos (θ) = 0

Expresar con seno, coseno

5 (\frac{1}{cos ( θ )})^{2} sin (θ) - cos (θ) = 0

Simplificar

5 (\frac{1}{cos ( θ )})^{2} sin (θ) - cos (θ) : \frac{5 sin ( θ ) - cos ^{3} ( θ )}{cos ^{2} ( θ )}

5 (\frac{1}{cos ( θ )})^{2} sin (θ) - cos (θ)

5 (\frac{1}{cos ( θ )})^{2} sin (θ) = \frac{5 sin ( θ )}{cos ^{2} ( θ )}

5 (\frac{1}{cos ( θ )})^{2} sin (θ)

(\frac{1}{cos ( θ )})^{2} = \frac{1}{cos ^{2} ( θ )}

(\frac{1}{cos ( θ )})^{2}

Aplicar las leyes de los exponentes:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{cos ^{2} ( θ )}

Aplicar la regla

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( θ )}

= 5 \cdot \frac{1}{cos ^{2} ( θ )} sin (θ)

Multiplicar fracciones:

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

= \frac{1 \cdot 5 sin ( θ )}{cos ^{2} ( θ )}

Multiplicar los numeros:

1 \cdot 5 = 5

= \frac{5 sin ( θ )}{cos ^{2} ( θ )}

= \frac{5 sin ( θ )}{cos ^{2} ( θ )} - cos (θ)

Convertir a fracción:

cos (θ) = \frac{cos ( θ ) cos ^{2} ( θ )}{cos ^{2} ( θ )}

= \frac{5 sin ( θ )}{cos ^{2} ( θ )} - \frac{cos ( θ ) cos ^{2} ( θ )}{cos ^{2} ( θ )}

Ya que los denominadores son iguales, combinar las fracciones:

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

= \frac{5 sin ( θ ) - cos ( θ ) cos ^{2} ( θ )}{cos ^{2} ( θ )}

5 sin (θ) - cos (θ) cos^{2} (θ) = 5 sin (θ) - cos^{3} (θ)

5 sin (θ) - cos (θ) cos^{2} (θ)

cos (θ) cos^{2} (θ) = cos^{3} (θ)

cos (θ) cos^{2} (θ)

Aplicar las leyes de los exponentes:

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

cos (θ) cos^{2} (θ) = cos^{1 + 2} (θ)

= cos^{1 + 2} (θ)

Sumar:

1 + 2 = 3

= cos^{3} (θ)

= 5 sin (θ) - cos^{3} (θ)

= \frac{5 sin ( θ ) - cos ^{3} ( θ )}{cos ^{2} ( θ )}

\frac{5 sin ( θ ) - cos ^{3} ( θ )}{cos ^{2} ( θ )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

5 sin (θ) - cos^{3} (θ) = 0

Sumar

cos^{3} (θ)

a ambos lados

5 sin (θ) = cos^{3} (θ)

Elevar al cuadrado ambos lados

(5 sin (θ))^{2} = (cos^{3} (θ))^{2}

Restar

(cos^{3} (θ))^{2}

de ambos lados

25 sin^{2} (θ) - cos^{6} (θ) = 0

Re-escribir usando identidades trigonométricas

- cos^{6} (θ) + 25 sin^{2} (θ)

Utilizar la identidad pitagórica:

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

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

= - cos^{6} (θ) + 25 (1 - cos^{2} (θ))

- cos^{6} (θ) + (1 - cos^{2} (θ)) \cdot 25 = 0

Usando el método de sustitución

- cos^{6} (θ) + (1 - cos^{2} (θ)) \cdot 25 = 0

Sea:

cos (θ) = u

- u^{6} + (1 - u^{2}) \cdot 25 = 0

- u^{6} + (1 - u^{2}) \cdot 25 = 0 : u = 0.96414 \dots, u = - 0.96414 \dots

- u^{6} + (1 - u^{2}) \cdot 25 = 0

Desarrollar

- u^{6} + (1 - u^{2}) \cdot 25 : - u^{6} + 25 - 25 u^{2}

- u^{6} + (1 - u^{2}) \cdot 25

= - u^{6} + 25 (1 - u^{2})

Expandir

25 (1 - u^{2}) : 25 - 25 u^{2}

25 (1 - u^{2})

Poner los parentesis utilizando:

a (b - c) = ab - a c

a = 25, b = 1, c = u^{2}

= 25 \cdot 1 - 25 u^{2}

Multiplicar los numeros:

25 \cdot 1 = 25

= 25 - 25 u^{2}

= - u^{6} + 25 - 25 u^{2}

- u^{6} + 25 - 25 u^{2} = 0

Escribir en la forma binómica

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

- u^{6} - 25 u^{2} + 25 = 0

Re-escribir la ecuación con

v = u^{2}

v^{3} = u^{6}

- v^{3} - 25 v + 25 = 0

Resolver

- v^{3} - 25 v + 25 = 0 : v \approx 0.96414 \dots

- v^{3} - 25 v + 25 = 0

Encontrar una solución para

- v^{3} - 25 v + 25 = 0

utilizando el método de Newton-Raphson:

v \approx 0.96414 \dots

- v^{3} - 25 v + 25 = 0

Definición del método de Newton-Raphson

f (v) = - v^{3} - 25 v + 25

Hallar

f^{^{'}} (v) : - 3 v^{2} - 25

\frac{d}{d v} (- v^{3} - 25 v + 25)

Aplicar la regla de la suma/diferencia:

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

= - \frac{d}{d v} (v^{3}) - \frac{d}{d v} (25 v) + \frac{d}{d v} (25)

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

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

Aplicar la regla de la potencia:

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

= 3 v^{3 - 1}

Simplificar

= 3 v^{2}

\frac{d}{d v} (25 v) = 25

\frac{d}{d v} (25 v)

Sacar la constante:

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

= 25 \frac{d v}{d v}

Aplicar la regla de derivación:

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

= 25 \cdot 1

Simplificar

= 25

\frac{d}{d v} (25) = 0

\frac{d}{d v} (25)

Derivada de una constante:

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

= 0

= - 3 v^{2} - 25 + 0

Simplificar

= - 3 v^{2} - 25

Sea

v_{0} = 1

Calcular

v_{n + 1}

hasta que

Δ v_{n + 1} < 0.000001

v_{1} = 0.96428 \dots : Δ v_{1} = 0.03571 \dots

f (v_{0}) = - 1^{3} - 25 \cdot 1 + 25 = - 1

f^{^{'}} (v_{0}) = - 3 \cdot 1^{2} - 25 = - 28

v_{1} = 0.96428 \dots

Δ v_{1} = ∣ 0.96428 \dots - 1 ∣ = 0.03571 \dots

Δ v_{1} = 0.03571 \dots

v_{2} = 0.96414 \dots : Δ v_{2} = 0.00013 \dots

f (v_{1}) = - 0.96428 \dots^{3} - 25 \cdot 0.96428 \dots + 25 = - 0.00378 \dots

f^{^{'}} (v_{1}) = - 3 \cdot 0.96428 \dots^{2} - 25 = - 27.78954 \dots

v_{2} = 0.96414 \dots

Δ v_{2} = ∣ 0.96414 \dots - 0.96428 \dots ∣ = 0.00013 \dots

Δ v_{2} = 0.00013 \dots

v_{3} = 0.96414 \dots : Δ v_{3} = 1.927 E - 9

f (v_{2}) = - 0.96414 \dots^{3} - 25 \cdot 0.96414 \dots + 25 = - 5.35491 E - 8

f^{^{'}} (v_{2}) = - 3 \cdot 0.96414 \dots^{2} - 25 = - 27.78875 \dots

v_{3} = 0.96414 \dots

Δ v_{3} = ∣ 0.96414 \dots - 0.96414 \dots ∣ = 1.927 E - 9

Δ v_{3} = 1.927 E - 9

v \approx 0.96414 \dots

Aplicar la división larga

Eq u a t i o n 0 : \frac{- v ^{3} - 25 v + 25}{v - 0.96414 \dots} = - v^{2} - 0.96414 \dots v - 25.92958 \dots

- v^{2} - 0.96414 \dots v - 25.92958 \dots \approx 0

Encontrar una solución para

- v^{2} - 0.96414 \dots v - 25.92958 \dots = 0

utilizando el método de Newton-Raphson:

Sin solución para

v \in R

- v^{2} - 0.96414 \dots v - 25.92958 \dots = 0

Definición del método de Newton-Raphson

f (v) = - v^{2} - 0.96414 \dots v - 25.92958 \dots

Hallar

f^{^{'}} (v) : - 2 v - 0.96414 \dots

\frac{d}{d v} (- v^{2} - 0.96414 \dots v - 25.92958 \dots)

Aplicar la regla de la suma/diferencia:

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

= - \frac{d}{d v} (v^{2}) - \frac{d}{d v} (0.96414 \dots v) - \frac{d}{d v} (25.92958 \dots)

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

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

Aplicar la regla de la potencia:

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

= 2 v^{2 - 1}

Simplificar

= 2 v

\frac{d}{d v} (0.96414 \dots v) = 0.96414 \dots

\frac{d}{d v} (0.96414 \dots v)

Sacar la constante:

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

= 0.96414 \dots \frac{d v}{d v}

Aplicar la regla de derivación:

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

= 0.96414 \dots \cdot 1

Simplificar

= 0.96414 \dots

\frac{d}{d v} (25.92958 \dots) = 0

\frac{d}{d v} (25.92958 \dots)

Derivada de una constante:

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

= 0

= - 2 v - 0.96414 \dots - 0

Simplificar

= - 2 v - 0.96414 \dots

Sea

v_{0} = - 5

Calcular

v_{n + 1}

hasta que

Δ v_{n + 1} < 0.000001

v_{1} = 0.10287 \dots : Δ v_{1} = 5.10287 \dots

f (v_{0}) = - (- 5)^{2} - 0.96414 \dots (- 5) - 25.92958 \dots = - 46.10883 \dots

f^{^{'}} (v_{0}) = - 2 (- 5) - 0.96414 \dots = 9.03585 \dots

v_{1} = 0.10287 \dots

Δ v_{1} = ∣ 0.10287 \dots - (- 5) ∣ = 5.10287 \dots

Δ v_{1} = 5.10287 \dots

v_{2} = - 22.15480 \dots : Δ v_{2} = 22.25767 \dots

f (v_{1}) = - 0.10287 \dots^{2} - 0.96414 \dots \cdot 0.10287 \dots - 25.92958 \dots = - 26.03935 \dots

f^{^{'}} (v_{1}) = - 2 \cdot 0.10287 \dots - 0.96414 \dots = - 1.16990 \dots

v_{2} = - 22.15480 \dots

Δ v_{2} = ∣ - 22.15480 \dots - 0.10287 \dots ∣ = 22.25767 \dots

Δ v_{2} = 22.25767 \dots

v_{3} = - 10.72559 \dots : Δ v_{3} = 11.42920 \dots

f (v_{2}) = - (- 22.15480 \dots)^{2} - 0.96414 \dots (- 22.15480 \dots) - 25.92958 \dots = - 495.40430 \dots

f^{^{'}} (v_{2}) = - 2 (- 22.15480 \dots) - 0.96414 \dots = 43.34545 \dots

v_{3} = - 10.72559 \dots

Δ v_{3} = ∣ - 10.72559 \dots - (- 22.15480 \dots) ∣ = 11.42920 \dots

Δ v_{3} = 11.42920 \dots

v_{4} = - 4.34951 \dots : Δ v_{4} = 6.37607 \dots

f (v_{3}) = - (- 10.72559 \dots)^{2} - 0.96414 \dots (- 10.72559 \dots) - 25.92958 \dots = - 130.62684 \dots

f^{^{'}} (v_{3}) = - 2 (- 10.72559 \dots) - 0.96414 \dots = 20.48703 \dots

v_{4} = - 4.34951 \dots

Δ v_{4} = ∣ - 4.34951 \dots - (- 10.72559 \dots) ∣ = 6.37607 \dots

Δ v_{4} = 6.37607 \dots

v_{5} = 0.90644 \dots : Δ v_{5} = 5.25596 \dots

f (v_{4}) = - (- 4.34951 \dots)^{2} - 0.96414 \dots (- 4.34951 \dots) - 25.92958 \dots = - 40.65431 \dots

f^{^{'}} (v_{4}) = - 2 (- 4.34951 \dots) - 0.96414 \dots = 7.73488 \dots

v_{5} = 0.90644 \dots

Δ v_{5} = ∣ 0.90644 \dots - (- 4.34951 \dots) ∣ = 5.25596 \dots

Δ v_{5} = 5.25596 \dots

v_{6} = - 9.04124 \dots : Δ v_{6} = 9.94769 \dots

f (v_{5}) = - 0.90644 \dots^{2} - 0.96414 \dots \cdot 0.90644 \dots - 25.92958 \dots = - 27.62518 \dots

f^{^{'}} (v_{5}) = - 2 \cdot 0.90644 \dots - 0.96414 \dots = - 2.77704 \dots

v_{6} = - 9.04124 \dots

Δ v_{6} = ∣ - 9.04124 \dots - 0.90644 \dots ∣ = 9.94769 \dots

Δ v_{6} = 9.94769 \dots

v_{7} = - 3.26050 \dots : Δ v_{7} = 5.78073 \dots

f (v_{6}) = - (- 9.04124 \dots)^{2} - 0.96414 \dots (- 9.04124 \dots) - 25.92958 \dots = - 98.95656 \dots

f^{^{'}} (v_{6}) = - 2 (- 9.04124 \dots) - 0.96414 \dots = 17.11833 \dots

v_{7} = - 3.26050 \dots

Δ v_{7} = ∣ - 3.26050 \dots - (- 9.04124 \dots) ∣ = 5.78073 \dots

Δ v_{7} = 5.78073 \dots

v_{8} = 2.75310 \dots : Δ v_{8} = 6.01361 \dots

f (v_{7}) = - (- 3.26050 \dots)^{2} - 0.96414 \dots (- 3.26050 \dots) - 25.92958 \dots = - 33.41688 \dots

f^{^{'}} (v_{7}) = - 2 (- 3.26050 \dots) - 0.96414 \dots = 5.55687 \dots

v_{8} = 2.75310 \dots

Δ v_{8} = ∣ 2.75310 \dots - (- 3.26050 \dots) ∣ = 6.01361 \dots

Δ v_{8} = 6.01361 \dots

v_{9} = - 2.83600 \dots : Δ v_{9} = 5.58911 \dots

f (v_{8}) = - 2.75310 \dots^{2} - 0.96414 \dots \cdot 2.75310 \dots - 25.92958 \dots = - 36.16359 \dots

f^{^{'}} (v_{8}) = - 2 \cdot 2.75310 \dots - 0.96414 \dots = - 6.47036 \dots

v_{9} = - 2.83600 \dots

Δ v_{9} = ∣ - 2.83600 \dots - 2.75310 \dots ∣ = 5.58911 \dots

Δ v_{9} = 5.58911 \dots

v_{10} = 3.79931 \dots : Δ v_{10} = 6.63532 \dots

f (v_{9}) = - (- 2.83600 \dots)^{2} - 0.96414 \dots (- 2.83600 \dots) - 25.92958 \dots = - 31.23817 \dots

f^{^{'}} (v_{9}) = - 2 (- 2.83600 \dots) - 0.96414 \dots = 4.70786 \dots

v_{10} = 3.79931 \dots

Δ v_{10} = ∣ 3.79931 \dots - (- 2.83600 \dots) ∣ = 6.63532 \dots

Δ v_{10} = 6.63532 \dots

No se puede encontrar solución

La solución es

v \approx 0.96414 \dots

v \approx 0.96414 \dots

Sustituir hacia atrás la

v = u^{2},

resolver para

u

Resolver

u^{2} = 0.96414 \dots : u = 0.96414 \dots, u = - 0.96414 \dots

u^{2} = 0.96414 \dots

Para

x^{2} = f (a)

las soluciones son

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

u = 0.96414 \dots, u = - 0.96414 \dots

Las soluciones son

u = 0.96414 \dots, u = - 0.96414 \dots

Sustituir en la ecuación

u = cos (θ)

cos (θ) = 0.96414 \dots, cos (θ) = - 0.96414 \dots

cos (θ) = 0.96414 \dots, cos (θ) = - 0.96414 \dots

cos (θ) = 0.96414 \dots : θ = arccos (0.96414 \dots) + 2 πn, θ = 2 π - arccos (0.96414 \dots) + 2 πn

cos (θ) = 0.96414 \dots

Aplicar propiedades trigonométricas inversas

cos (θ) = 0.96414 \dots

Soluciones generales para

cos (θ) = 0.96414 \dots

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

θ = arccos (0.96414 \dots) + 2 πn, θ = 2 π - arccos (0.96414 \dots) + 2 πn

θ = arccos (0.96414 \dots) + 2 πn, θ = 2 π - arccos (0.96414 \dots) + 2 πn

cos (θ) = - 0.96414 \dots : θ = arccos (- 0.96414 \dots) + 2 πn, θ = - arccos (- 0.96414 \dots) + 2 πn

cos (θ) = - 0.96414 \dots

Aplicar propiedades trigonométricas inversas

cos (θ) = - 0.96414 \dots

Soluciones generales para

cos (θ) = - 0.96414 \dots

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

θ = arccos (- 0.96414 \dots) + 2 πn, θ = - arccos (- 0.96414 \dots) + 2 πn

θ = arccos (- 0.96414 \dots) + 2 πn, θ = - arccos (- 0.96414 \dots) + 2 πn

Combinar toda las soluciones

θ = arccos (0.96414 \dots) + 2 πn, θ = 2 π - arccos (0.96414 \dots) + 2 πn, θ = arccos (- 0.96414 \dots) + 2 πn, θ = - arccos (- 0.96414 \dots) + 2 πn

Verificar las soluciones sustituyendo en la ecuación original

Verificar las soluciones sustituyéndolas en

5 sec^{2} (θ) sin (θ) - cos (θ) = 0

Quitar las que no concuerden con la ecuación.

Verificar la solución

arccos (0.96414 \dots) + 2 πn :

Verdadero

arccos (0.96414 \dots) + 2 πn

Sustituir

n = 1

arccos (0.96414 \dots) + 2 π 1

Multiplicar

5 sec^{2} (θ) sin (θ) - cos (θ) = 0

por

θ = arccos (0.96414 \dots) + 2 π 1

5 sec^{2} (arccos (0.96414 \dots) + 2 π 1) sin (arccos (0.96414 \dots) + 2 π 1) - cos (arccos (0.96414 \dots) + 2 π 1) = 0

Simplificar

0 = 0

\Rightarrow Verdadero

Verificar la solución

2 π - arccos (0.96414 \dots) + 2 πn :

Falso

2 π - arccos (0.96414 \dots) + 2 πn

Sustituir

n = 1

2 π - arccos (0.96414 \dots) + 2 π 1

Multiplicar

5 sec^{2} (θ) sin (θ) - cos (θ) = 0

por

θ = 2 π - arccos (0.96414 \dots) + 2 π 1

5 sec^{2} (2 π - arccos (0.96414 \dots) + 2 π 1) sin (2 π - arccos (0.96414 \dots) + 2 π 1) - cos (2 π - arccos (0.96414 \dots) + 2 π 1) = 0

Simplificar

- 1.96382 \dots = 0

\Rightarrow Falso

Verificar la solución

arccos (- 0.96414 \dots) + 2 πn :

Falso

arccos (- 0.96414 \dots) + 2 πn

Sustituir

n = 1

arccos (- 0.96414 \dots) + 2 π 1

Multiplicar

5 sec^{2} (θ) sin (θ) - cos (θ) = 0

por

θ = arccos (- 0.96414 \dots) + 2 π 1

5 sec^{2} (arccos (- 0.96414 \dots) + 2 π 1) sin (arccos (- 0.96414 \dots) + 2 π 1) - cos (arccos (- 0.96414 \dots) + 2 π 1) = 0

Simplificar

1.96382 \dots = 0

\Rightarrow Falso

Verificar la solución

- arccos (- 0.96414 \dots) + 2 πn :

Verdadero

- arccos (- 0.96414 \dots) + 2 πn

Sustituir

n = 1

- arccos (- 0.96414 \dots) + 2 π 1

Multiplicar

5 sec^{2} (θ) sin (θ) - cos (θ) = 0

por

θ = - arccos (- 0.96414 \dots) + 2 π 1

5 sec^{2} (- arccos (- 0.96414 \dots) + 2 π 1) sin (- arccos (- 0.96414 \dots) + 2 π 1) - cos (- arccos (- 0.96414 \dots) + 2 π 1) = 0

Simplificar

0 = 0

\Rightarrow Verdadero

θ = arccos (0.96414 \dots) + 2 πn, θ = - arccos (- 0.96414 \dots) + 2 πn

Mostrar soluciones en forma decimal

θ = 0.19049 \dots + 2 πn, θ = - 2.95110 \dots + 2 πn

Ejemplos populares

cos(x)= 8/11

cos (x) = \frac{8}{11}

4sec(θ)-sqrt(3)=sqrt(3)+7sec(θ)

4 sec (θ) - 3 = 3 + 7 sec (θ)

1/(sqrt(15))=cos(x)

\frac{1}{15} = cos (x)

cos^2(θ)=3sin^2(θ)

cos^{2} (θ) = 3 sin^{2} (θ)

sinh(x)= 5/4

sinh (x) = \frac{5}{4}