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csc^3(x)-4csc(x)=3cot^2(x)

Solución

csc^{3} (x) - 4 csc (x) = 3 cot^{2} (x)

Solución

x = - 0.79731 \dots + 2 πn, x = π + 0.79731 \dots + 2 πn, x = 0.26356 \dots + 2 πn, x = π - 0.26356 \dots + 2 πn

Grados

x = - 45.68261 \dots^{\circ} + 36 0^{\circ} n, x = 225.68261 \dots^{\circ} + 36 0^{\circ} n, x = 15.10095 \dots^{\circ} + 36 0^{\circ} n, x = 164.89904 \dots^{\circ} + 36 0^{\circ} n

Pasos de solución

csc^{3} (x) - 4 csc (x) = 3 cot^{2} (x)

Restar

3 cot^{2} (x)

de ambos lados

csc^{3} (x) - 4 csc (x) - 3 cot^{2} (x) = 0

Re-escribir usando identidades trigonométricas

csc^{3} (x) - 3 cot^{2} (x) - 4 csc (x)

Utilizar la identidad pitagórica:

1 + cot^{2} (x) = csc^{2} (x)

cot^{2} (x) = csc^{2} (x) - 1

= csc^{3} (x) - 3 (csc^{2} (x) - 1) - 4 csc (x)

csc^{3} (x) - (- 1 + csc^{2} (x)) \cdot 3 - 4 csc (x) = 0

Usando el método de sustitución

csc^{3} (x) - (- 1 + csc^{2} (x)) \cdot 3 - 4 csc (x) = 0

Sea:

csc (x) = u

u^{3} - (- 1 + u^{2}) \cdot 3 - 4 u = 0

u^{3} - (- 1 + u^{2}) \cdot 3 - 4 u = 0 : u \approx 0.55919 \dots, u \approx - 1.39766 \dots, u \approx 3.83846 \dots

u^{3} - (- 1 + u^{2}) \cdot 3 - 4 u = 0

Desarrollar

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

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

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

Expandir

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

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

Poner los parentesis utilizando:

a (b + c) = ab + a c

a = - 3, b = - 1, c = u^{2}

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

Aplicar las reglas de los signos

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

= 3 \cdot 1 - 3 u^{2}

Multiplicar los numeros:

3 \cdot 1 = 3

= 3 - 3 u^{2}

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

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

Escribir en la forma binómica

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

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

Encontrar una solución para

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

utilizando el método de Newton-Raphson:

u \approx 0.55919 \dots

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

Definición del método de Newton-Raphson

f (u) = u^{3} - 3 u^{2} - 4 u + 3

Hallar

f^{^{'}} (u) : 3 u^{2} - 6 u - 4

\frac{d}{d u} (u^{3} - 3 u^{2} - 4 u + 3)

Aplicar la regla de la suma/diferencia:

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

= \frac{d}{d u} (u^{3}) - \frac{d}{d u} (3 u^{2}) - \frac{d}{d u} (4 u) + \frac{d}{d u} (3)

\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} (3 u^{2}) = 6 u

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

Sacar la constante:

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

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

Aplicar la regla de la potencia:

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

= 3 \cdot 2 u^{2 - 1}

Simplificar

= 6 u

\frac{d}{d u} (4 u) = 4

\frac{d}{d u} (4 u)

Sacar la constante:

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

= 4 \frac{d u}{d u}

Aplicar la regla de derivación:

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

= 4 \cdot 1

Simplificar

= 4

\frac{d}{d u} (3) = 0

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

Derivada de una constante:

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

= 0

= 3 u^{2} - 6 u - 4 + 0

Simplificar

= 3 u^{2} - 6 u - 4

Sea

u_{0} = 1

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = 0.57142 \dots : Δ u_{1} = 0.42857 \dots

f (u_{0}) = 1^{3} - 3 \cdot 1^{2} - 4 \cdot 1 + 3 = - 3

f^{^{'}} (u_{0}) = 3 \cdot 1^{2} - 6 \cdot 1 - 4 = - 7

u_{1} = 0.57142 \dots

Δ u_{1} = ∣ 0.57142 \dots - 1 ∣ = 0.42857 \dots

Δ u_{1} = 0.42857 \dots

u_{2} = 0.55922 \dots : Δ u_{2} = 0.01220 \dots

f (u_{1}) = 0.57142 \dots^{3} - 3 \cdot 0.57142 \dots^{2} - 4 \cdot 0.57142 \dots + 3 = - 0.07871 \dots

f^{^{'}} (u_{1}) = 3 \cdot 0.57142 \dots^{2} - 6 \cdot 0.57142 \dots - 4 = - 6.44897 \dots

u_{2} = 0.55922 \dots

Δ u_{2} = ∣ 0.55922 \dots - 0.57142 \dots ∣ = 0.01220 \dots

Δ u_{2} = 0.01220 \dots

u_{3} = 0.55919 \dots : Δ u_{3} = 0.00003 \dots

f (u_{2}) = 0.55922 \dots^{3} - 3 \cdot 0.55922 \dots^{2} - 4 \cdot 0.55922 \dots + 3 = - 0.00019 \dots

f^{^{'}} (u_{2}) = 3 \cdot 0.55922 \dots^{2} - 6 \cdot 0.55922 \dots - 4 = - 6.41714 \dots

u_{3} = 0.55919 \dots

Δ u_{3} = ∣ 0.55919 \dots - 0.55922 \dots ∣ = 0.00003 \dots

Δ u_{3} = 0.00003 \dots

u_{4} = 0.55919 \dots : Δ u_{4} = 1.87129 E - 10

f (u_{3}) = 0.55919 \dots^{3} - 3 \cdot 0.55919 \dots^{2} - 4 \cdot 0.55919 \dots + 3 = - 1.20082 E - 9

f^{^{'}} (u_{3}) = 3 \cdot 0.55919 \dots^{2} - 6 \cdot 0.55919 \dots - 4 = - 6.41706 \dots

u_{4} = 0.55919 \dots

Δ u_{4} = ∣ 0.55919 \dots - 0.55919 \dots ∣ = 1.87129 E - 10

Δ u_{4} = 1.87129 E - 10

u \approx 0.55919 \dots

Aplicar la división larga

Eq u a t i o n 0 : \frac{u ^{3} - 3 u ^{2} - 4 u + 3}{u - 0.55919 \dots} = u^{2} - 2.44080 \dots u - 5.36488 \dots

u^{2} - 2.44080 \dots u - 5.36488 \dots \approx 0

Encontrar una solución para

u^{2} - 2.44080 \dots u - 5.36488 \dots = 0

utilizando el método de Newton-Raphson:

u \approx - 1.39766 \dots

u^{2} - 2.44080 \dots u - 5.36488 \dots = 0

Definición del método de Newton-Raphson

f (u) = u^{2} - 2.44080 \dots u - 5.36488 \dots

Hallar

f^{^{'}} (u) : 2 u - 2.44080 \dots

\frac{d}{d u} (u^{2} - 2.44080 \dots u - 5.36488 \dots)

Aplicar la regla de la suma/diferencia:

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

= \frac{d}{d u} (u^{2}) - \frac{d}{d u} (2.44080 \dots u) - \frac{d}{d u} (5.36488 \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} (2.44080 \dots u) = 2.44080 \dots

\frac{d}{d u} (2.44080 \dots u)

Sacar la constante:

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

= 2.44080 \dots \frac{d u}{d u}

Aplicar la regla de derivación:

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

= 2.44080 \dots \cdot 1

Simplificar

= 2.44080 \dots

\frac{d}{d u} (5.36488 \dots) = 0

\frac{d}{d u} (5.36488 \dots)

Derivada de una constante:

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

= 0

= 2 u - 2.44080 \dots - 0

Simplificar

= 2 u - 2.44080 \dots

Sea

u_{0} = - 2

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = - 1.45399 \dots : Δ u_{1} = 0.54600 \dots

f (u_{0}) = (- 2)^{2} - 2.44080 \dots (- 2) - 5.36488 \dots = 3.51673 \dots

f^{^{'}} (u_{0}) = 2 (- 2) - 2.44080 \dots = - 6.44080 \dots

u_{1} = - 1.45399 \dots

Δ u_{1} = ∣ - 1.45399 \dots - (- 2) ∣ = 0.54600 \dots

Δ u_{1} = 0.54600 \dots

u_{2} = - 1.39825 \dots : Δ u_{2} = 0.05573 \dots

f (u_{1}) = (- 1.45399 \dots)^{2} - 2.44080 \dots (- 1.45399 \dots) - 5.36488 \dots = 0.29812 \dots

f^{^{'}} (u_{1}) = 2 (- 1.45399 \dots) - 2.44080 \dots = - 5.34879 \dots

u_{2} = - 1.39825 \dots

Δ u_{2} = ∣ - 1.39825 \dots - (- 1.45399 \dots) ∣ = 0.05573 \dots

Δ u_{2} = 0.05573 \dots

u_{3} = - 1.39766 \dots : Δ u_{3} = 0.00059 \dots

f (u_{2}) = (- 1.39825 \dots)^{2} - 2.44080 \dots (- 1.39825 \dots) - 5.36488 \dots = 0.00310 \dots

f^{^{'}} (u_{2}) = 2 (- 1.39825 \dots) - 2.44080 \dots = - 5.23731 \dots

u_{3} = - 1.39766 \dots

Δ u_{3} = ∣ - 1.39766 \dots - (- 1.39825 \dots) ∣ = 0.00059 \dots

Δ u_{3} = 0.00059 \dots

u_{4} = - 1.39766 \dots : Δ u_{4} = 6.7196 E - 8

f (u_{3}) = (- 1.39766 \dots)^{2} - 2.44080 \dots (- 1.39766 \dots) - 5.36488 \dots = 3.51847 E - 7

f^{^{'}} (u_{3}) = 2 (- 1.39766 \dots) - 2.44080 \dots = - 5.23613 \dots

u_{4} = - 1.39766 \dots

Δ u_{4} = ∣ - 1.39766 \dots - (- 1.39766 \dots) ∣ = 6.7196 E - 8

Δ u_{4} = 6.7196 E - 8

u \approx - 1.39766 \dots

Aplicar la división larga

Eq u a t i o n 0 : \frac{u ^{2} - 2.44080 \dots u - 5.36488 \dots}{u + 1.39766 \dots} = u - 3.83846 \dots

u - 3.83846 \dots \approx 0

u \approx 3.83846 \dots

Las soluciones son

u \approx 0.55919 \dots, u \approx - 1.39766 \dots, u \approx 3.83846 \dots

Sustituir en la ecuación

u = csc (x)

csc (x) \approx 0.55919 \dots, csc (x) \approx - 1.39766 \dots, csc (x) \approx 3.83846 \dots

csc (x) \approx 0.55919 \dots, csc (x) \approx - 1.39766 \dots, csc (x) \approx 3.83846 \dots

csc (x) = 0.55919 \dots :

Sin solución

csc (x) = 0.55919 \dots

csc (x) \leq - 1 or csc (x) \geq 1

Sin soluci \overset{o}{ˊ} n

csc (x) = - 1.39766 \dots : x = arccsc (- 1.39766 \dots) + 2 πn, x = π + arccsc (1.39766 \dots) + 2 πn

csc (x) = - 1.39766 \dots

Aplicar propiedades trigonométricas inversas

csc (x) = - 1.39766 \dots

Soluciones generales para

csc (x) = - 1.39766 \dots

csc (x) = - a \Rightarrow x = arccsc (- a) + 2 πn, x = π + arccsc (a) + 2 πn

x = arccsc (- 1.39766 \dots) + 2 πn, x = π + arccsc (1.39766 \dots) + 2 πn

x = arccsc (- 1.39766 \dots) + 2 πn, x = π + arccsc (1.39766 \dots) + 2 πn

csc (x) = 3.83846 \dots : x = arccsc (3.83846 \dots) + 2 πn, x = π - arccsc (3.83846 \dots) + 2 πn

csc (x) = 3.83846 \dots

Aplicar propiedades trigonométricas inversas

csc (x) = 3.83846 \dots

Soluciones generales para

csc (x) = 3.83846 \dots

csc (x) = a \Rightarrow x = arccsc (a) + 2 πn, x = π - arccsc (a) + 2 πn

x = arccsc (3.83846 \dots) + 2 πn, x = π - arccsc (3.83846 \dots) + 2 πn

x = arccsc (3.83846 \dots) + 2 πn, x = π - arccsc (3.83846 \dots) + 2 πn

Combinar toda las soluciones

x = arccsc (- 1.39766 \dots) + 2 πn, x = π + arccsc (1.39766 \dots) + 2 πn, x = arccsc (3.83846 \dots) + 2 πn, x = π - arccsc (3.83846 \dots) + 2 πn

Mostrar soluciones en forma decimal

x = - 0.79731 \dots + 2 πn, x = π + 0.79731 \dots + 2 πn, x = 0.26356 \dots + 2 πn, x = π - 0.26356 \dots + 2 πn

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