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50sec^2(5x)tan(5x)=25+tan^2(5x)

Solución

50 sec^{2} (5 x) tan (5 x) = 25 + tan^{2} (5 x)

Solución

x = \frac{0.40289 \dots}{5} + \frac{πn}{5}

Grados

x = 4.61683 \dots^{\circ} + 3 6^{\circ} n

Pasos de solución

50 sec^{2} (5 x) tan (5 x) = 25 + tan^{2} (5 x)

Restar

25 + tan^{2} (5 x)

de ambos lados

50 sec^{2} (5 x) tan (5 x) - 25 - tan^{2} (5 x) = 0

Re-escribir usando identidades trigonométricas

- 25 - tan^{2} (5 x) + 50 sec^{2} (5 x) tan (5 x)

Utilizar la identidad pitagórica:

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

= - 25 - tan^{2} (5 x) + 50 (tan^{2} (5 x) + 1) tan (5 x)

- 25 - tan^{2} (5 x) + (1 + tan^{2} (5 x)) \cdot 50 tan (5 x) = 0

Usando el método de sustitución

- 25 - tan^{2} (5 x) + (1 + tan^{2} (5 x)) \cdot 50 tan (5 x) = 0

Sea:

tan (5 x) = u

- 25 - u^{2} + (1 + u^{2}) \cdot 50 u = 0

- 25 - u^{2} + (1 + u^{2}) \cdot 50 u = 0 : u \approx 0.42620 \dots

- 25 - u^{2} + (1 + u^{2}) \cdot 50 u = 0

Desarrollar

- 25 - u^{2} + (1 + u^{2}) \cdot 50 u : - 25 - u^{2} + 50 u + 50 u^{3}

- 25 - u^{2} + (1 + u^{2}) \cdot 50 u

= - 25 - u^{2} + 50 u (1 + u^{2})

Expandir

50 u (1 + u^{2}) : 50 u + 50 u^{3}

50 u (1 + u^{2})

Poner los parentesis utilizando:

a (b + c) = ab + a c

a = 50 u, b = 1, c = u^{2}

= 50 u \cdot 1 + 50 u u^{2}

= 50 \cdot 1 \cdot u + 50 u^{2} u

Simplificar

50 \cdot 1 \cdot u + 50 u^{2} u : 50 u + 50 u^{3}

50 \cdot 1 \cdot u + 50 u^{2} u

50 \cdot 1 \cdot u = 50 u

50 \cdot 1 \cdot u

Multiplicar los numeros:

50 \cdot 1 = 50

= 50 u

50 u^{2} u = 50 u^{3}

50 u^{2} u

Aplicar las leyes de los exponentes:

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

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

= 50 u^{2 + 1}

Sumar:

2 + 1 = 3

= 50 u^{3}

= 50 u + 50 u^{3}

= 50 u + 50 u^{3}

= - 25 - u^{2} + 50 u + 50 u^{3}

- 25 - u^{2} + 50 u + 50 u^{3} = 0

Escribir en la forma binómica

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

50 u^{3} - u^{2} + 50 u - 25 = 0

Encontrar una solución para

50 u^{3} - u^{2} + 50 u - 25 = 0

utilizando el método de Newton-Raphson:

u \approx 0.42620 \dots

50 u^{3} - u^{2} + 50 u - 25 = 0

Definición del método de Newton-Raphson

f (u) = 50 u^{3} - u^{2} + 50 u - 25

Hallar

f^{^{'}} (u) : 150 u^{2} - 2 u + 50

\frac{d}{d u} (50 u^{3} - u^{2} + 50 u - 25)

Aplicar la regla de la suma/diferencia:

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

= \frac{d}{d u} (50 u^{3}) - \frac{d}{d u} (u^{2}) + \frac{d}{d u} (50 u) - \frac{d}{d u} (25)

\frac{d}{d u} (50 u^{3}) = 150 u^{2}

\frac{d}{d u} (50 u^{3})

Sacar la constante:

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

= 50 \frac{d}{d u} (u^{3})

Aplicar la regla de la potencia:

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

= 50 \cdot 3 u^{3 - 1}

Simplificar

= 150 u^{2}

\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} (50 u) = 50

\frac{d}{d u} (50 u)

Sacar la constante:

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

= 50 \frac{d u}{d u}

Aplicar la regla de derivación:

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

= 50 \cdot 1

Simplificar

= 50

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

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

Derivada de una constante:

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

= 0

= 150 u^{2} - 2 u + 50 - 0

Simplificar

= 150 u^{2} - 2 u + 50

Sea

u_{0} = 1

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = 0.62626 \dots : Δ u_{1} = 0.37373 \dots

f (u_{0}) = 50 \cdot 1^{3} - 1^{2} + 50 \cdot 1 - 25 = 74

f^{^{'}} (u_{0}) = 150 \cdot 1^{2} - 2 \cdot 1 + 50 = 198

u_{1} = 0.62626 \dots

Δ u_{1} = ∣ 0.62626 \dots - 1 ∣ = 0.37373 \dots

Δ u_{1} = 0.37373 \dots

u_{2} = 0.45706 \dots : Δ u_{2} = 0.16919 \dots

f (u_{1}) = 50 \cdot 0.62626 \dots^{3} - 0.62626 \dots^{2} + 50 \cdot 0.62626 \dots - 25 = 18.20208 \dots

f^{^{'}} (u_{1}) = 150 \cdot 0.62626 \dots^{2} - 2 \cdot 0.62626 \dots + 50 = 107.57820 \dots

u_{2} = 0.45706 \dots

Δ u_{2} = ∣ 0.45706 \dots - 0.62626 \dots ∣ = 0.16919 \dots

Δ u_{2} = 0.16919 \dots

u_{3} = 0.42699 \dots : Δ u_{3} = 0.03007 \dots

f (u_{2}) = 50 \cdot 0.45706 \dots^{3} - 0.45706 \dots^{2} + 50 \cdot 0.45706 \dots - 25 = 2.41849 \dots

f^{^{'}} (u_{2}) = 150 \cdot 0.45706 \dots^{2} - 2 \cdot 0.45706 \dots + 50 = 80.42199 \dots

u_{3} = 0.42699 \dots

Δ u_{3} = ∣ 0.42699 \dots - 0.45706 \dots ∣ = 0.03007 \dots

Δ u_{3} = 0.03007 \dots

u_{4} = 0.42621 \dots : Δ u_{4} = 0.00078 \dots

f (u_{3}) = 50 \cdot 0.42699 \dots^{3} - 0.42699 \dots^{2} + 50 \cdot 0.42699 \dots - 25 = 0.05973 \dots

f^{^{'}} (u_{3}) = 150 \cdot 0.42699 \dots^{2} - 2 \cdot 0.42699 \dots + 50 = 76.49426 \dots

u_{4} = 0.42621 \dots

Δ u_{4} = ∣ 0.42621 \dots - 0.42699 \dots ∣ = 0.00078 \dots

Δ u_{4} = 0.00078 \dots

u_{5} = 0.42620 \dots : Δ u_{5} = 5.03019 E - 7

f (u_{4}) = 50 \cdot 0.42621 \dots^{3} - 0.42621 \dots^{2} + 50 \cdot 0.42621 \dots - 25 = 0.00003 \dots

f^{^{'}} (u_{4}) = 150 \cdot 0.42621 \dots^{2} - 2 \cdot 0.42621 \dots + 50 = 76.39588 \dots

u_{5} = 0.42620 \dots

Δ u_{5} = ∣ 0.42620 \dots - 0.42621 \dots ∣ = 5.03019 E - 7

Δ u_{5} = 5.03019 E - 7

u \approx 0.42620 \dots

Aplicar la división larga

Eq u a t i o n 0 : \frac{50 u ^{3} - u ^{2} + 50 u - 25}{u - 0.42620 \dots} = 50 u^{2} + 20.31049 \dots u + 58.65653 \dots

50 u^{2} + 20.31049 \dots u + 58.65653 \dots \approx 0

Encontrar una solución para

50 u^{2} + 20.31049 \dots u + 58.65653 \dots = 0

utilizando el método de Newton-Raphson:

Sin solución para

u \in R

50 u^{2} + 20.31049 \dots u + 58.65653 \dots = 0

Definición del método de Newton-Raphson

f (u) = 50 u^{2} + 20.31049 \dots u + 58.65653 \dots

Hallar

f^{^{'}} (u) : 100 u + 20.31049 \dots

\frac{d}{d u} (50 u^{2} + 20.31049 \dots u + 58.65653 \dots)

Aplicar la regla de la suma/diferencia:

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

= \frac{d}{d u} (50 u^{2}) + \frac{d}{d u} (20.31049 \dots u) + \frac{d}{d u} (58.65653 \dots)

\frac{d}{d u} (50 u^{2}) = 100 u

\frac{d}{d u} (50 u^{2})

Sacar la constante:

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

= 50 \frac{d}{d u} (u^{2})

Aplicar la regla de la potencia:

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

= 50 \cdot 2 u^{2 - 1}

Simplificar

= 100 u

\frac{d}{d u} (20.31049 \dots u) = 20.31049 \dots

\frac{d}{d u} (20.31049 \dots u)

Sacar la constante:

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

= 20.31049 \dots \frac{d u}{d u}

Aplicar la regla de derivación:

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

= 20.31049 \dots \cdot 1

Simplificar

= 20.31049 \dots

\frac{d}{d u} (58.65653 \dots) = 0

\frac{d}{d u} (58.65653 \dots)

Derivada de una constante:

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

= 0

= 100 u + 20.31049 \dots + 0

Simplificar

= 100 u + 20.31049 \dots

Sea

u_{0} = - 3

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = - 1.39920 \dots : Δ u_{1} = 1.60079 \dots

f (u_{0}) = 50 (- 3)^{2} + 20.31049 \dots (- 3) + 58.65653 \dots = 447.72504 \dots

f^{^{'}} (u_{0}) = 100 (- 3) + 20.31049 \dots = - 279.68950 \dots

u_{1} = - 1.39920 \dots

Δ u_{1} = ∣ - 1.39920 \dots - (- 3) ∣ = 1.60079 \dots

Δ u_{1} = 1.60079 \dots

u_{2} = - 0.32800 \dots : Δ u_{2} = 1.07120 \dots

f (u_{1}) = 50 (- 1.39920 \dots)^{2} + 20.31049 \dots (- 1.39920 \dots) + 58.65653 \dots = 128.12693 \dots

f^{^{'}} (u_{1}) = 100 (- 1.39920 \dots) + 20.31049 \dots = - 119.61018 \dots

u_{2} = - 0.32800 \dots

Δ u_{2} = ∣ - 0.32800 \dots - (- 1.39920 \dots) ∣ = 1.07120 \dots

Δ u_{2} = 1.07120 \dots

u_{3} = 4.26567 \dots : Δ u_{3} = 4.59367 \dots

f (u_{2}) = 50 (- 0.32800 \dots)^{2} + 20.31049 \dots (- 0.32800 \dots) + 58.65653 \dots = 57.37392 \dots

f^{^{'}} (u_{2}) = 100 (- 0.32800 \dots) + 20.31049 \dots = - 12.48976 \dots

u_{3} = 4.26567 \dots

Δ u_{3} = ∣ 4.26567 \dots - (- 0.32800 \dots) ∣ = 4.59367 \dots

Δ u_{3} = 4.59367 \dots

u_{4} = 1.90464 \dots : Δ u_{4} = 2.36103 \dots

f (u_{3}) = 50 \cdot 4.26567 \dots^{2} + 20.31049 \dots \cdot 4.26567 \dots + 58.65653 \dots = 1055.09312 \dots

f^{^{'}} (u_{3}) = 100 \cdot 4.26567 \dots + 20.31049 \dots = 446.87787 \dots

u_{4} = 1.90464 \dots

Δ u_{4} = ∣ 1.90464 \dots - 4.26567 \dots ∣ = 2.36103 \dots

Δ u_{4} = 2.36103 \dots

u_{5} = 0.58226 \dots : Δ u_{5} = 1.32237 \dots

f (u_{4}) = 50 \cdot 1.90464 \dots^{2} + 20.31049 \dots \cdot 1.90464 \dots + 58.65653 \dots = 278.72369 \dots

f^{^{'}} (u_{4}) = 100 \cdot 1.90464 \dots + 20.31049 \dots = 210.77463 \dots

u_{5} = 0.58226 \dots

Δ u_{5} = ∣ 0.58226 \dots - 1.90464 \dots ∣ = 1.32237 \dots

Δ u_{5} = 1.32237 \dots

u_{6} = - 0.53102 \dots : Δ u_{6} = 1.11328 \dots

f (u_{5}) = 50 \cdot 0.58226 \dots^{2} + 20.31049 \dots \cdot 0.58226 \dots + 58.65653 \dots = 87.43414 \dots

f^{^{'}} (u_{5}) = 100 \cdot 0.58226 \dots + 20.31049 \dots = 78.53686 \dots

u_{6} = - 0.53102 \dots

Δ u_{6} = ∣ - 0.53102 \dots - 0.58226 \dots ∣ = 1.11328 \dots

Δ u_{6} = 1.11328 \dots

u_{7} = 1.35878 \dots : Δ u_{7} = 1.88980 \dots

f (u_{6}) = 50 (- 0.53102 \dots)^{2} + 20.31049 \dots (- 0.53102 \dots) + 58.65653 \dots = 61.97051 \dots

f^{^{'}} (u_{6}) = 100 (- 0.53102 \dots) + 20.31049 \dots = - 32.79194 \dots

u_{7} = 1.35878 \dots

Δ u_{7} = ∣ 1.35878 \dots - (- 0.53102 \dots) ∣ = 1.88980 \dots

Δ u_{7} = 1.88980 \dots

u_{8} = 0.21549 \dots : Δ u_{8} = 1.14328 \dots

f (u_{7}) = 50 \cdot 1.35878 \dots^{2} + 20.31049 \dots \cdot 1.35878 \dots + 58.65653 \dots = 178.56892 \dots

f^{^{'}} (u_{7}) = 100 \cdot 1.35878 \dots + 20.31049 \dots = 156.18896 \dots

u_{8} = 0.21549 \dots

Δ u_{8} = ∣ 0.21549 \dots - 1.35878 \dots ∣ = 1.14328 \dots

Δ u_{8} = 1.14328 \dots

u_{9} = - 1.34577 \dots : Δ u_{9} = 1.56127 \dots

f (u_{8}) = 50 \cdot 0.21549 \dots^{2} + 20.31049 \dots \cdot 0.21549 \dots + 58.65653 \dots = 65.35533 \dots

f^{^{'}} (u_{8}) = 100 \cdot 0.21549 \dots + 20.31049 \dots = 41.86020 \dots

u_{9} = - 1.34577 \dots

Δ u_{9} = ∣ - 1.34577 \dots - 0.21549 \dots ∣ = 1.56127 \dots

Δ u_{9} = 1.56127 \dots

u_{10} = - 0.27916 \dots : Δ u_{10} = 1.06661 \dots

f (u_{9}) = 50 (- 1.34577 \dots)^{2} + 20.31049 \dots (- 1.34577 \dots) + 58.65653 \dots = 121.87917 \dots

f^{^{'}} (u_{9}) = 100 (- 1.34577 \dots) + 20.31049 \dots = - 114.26742 \dots

u_{10} = - 0.27916 \dots

Δ u_{10} = ∣ - 0.27916 \dots - (- 1.34577 \dots) ∣ = 1.06661 \dots

Δ u_{10} = 1.06661 \dots

u_{11} = 7.19949 \dots : Δ u_{11} = 7.47865 \dots

f (u_{10}) = 50 (- 0.27916 \dots)^{2} + 20.31049 \dots (- 0.27916 \dots) + 58.65653 \dots = 56.88321 \dots

f^{^{'}} (u_{10}) = 100 (- 0.27916 \dots) + 20.31049 \dots = - 7.60607 \dots

u_{11} = 7.19949 \dots

Δ u_{11} = ∣ 7.19949 \dots - (- 0.27916 \dots) ∣ = 7.47865 \dots

Δ u_{11} = 7.47865 \dots

No se puede encontrar solución

La solución es

u \approx 0.42620 \dots

Sustituir en la ecuación

u = tan (5 x)

tan (5 x) \approx 0.42620 \dots

tan (5 x) \approx 0.42620 \dots

tan (5 x) = 0.42620 \dots : x = \frac{arctan ( 0.42620 \dots )}{5} + \frac{πn}{5}

tan (5 x) = 0.42620 \dots

Aplicar propiedades trigonométricas inversas

tan (5 x) = 0.42620 \dots

Soluciones generales para

tan (5 x) = 0.42620 \dots

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

5 x = arctan (0.42620 \dots) + πn

5 x = arctan (0.42620 \dots) + πn

Resolver

5 x = arctan (0.42620 \dots) + πn : x = \frac{arctan ( 0.42620 \dots )}{5} + \frac{πn}{5}

5 x = arctan (0.42620 \dots) + πn

Dividir ambos lados entre

5

5 x = arctan (0.42620 \dots) + πn

Dividir ambos lados entre

5

\frac{5 x}{5} = \frac{arctan ( 0.42620 \dots )}{5} + \frac{πn}{5}

Simplificar

x = \frac{arctan ( 0.42620 \dots )}{5} + \frac{πn}{5}

x = \frac{arctan ( 0.42620 \dots )}{5} + \frac{πn}{5}

x = \frac{arctan ( 0.42620 \dots )}{5} + \frac{πn}{5}

Combinar toda las soluciones

x = \frac{arctan ( 0.42620 \dots )}{5} + \frac{πn}{5}

Mostrar soluciones en forma decimal

x = \frac{0.40289 \dots}{5} + \frac{πn}{5}

Ejemplos populares

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