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Popular Trigonometría >

sin^3(x)-2sin(x)=4cos^2(x)-3

Solución

sin^{3} (x) - 2 sin (x) = 4 cos^{2} (x) - 3

Solución

x = - 0.32154 \dots + 2 πn, x = π + 0.32154 \dots + 2 πn, x = 0.80194 \dots + 2 πn, x = π - 0.80194 \dots + 2 πn

Grados

x = - 18.42305 \dots^{\circ} + 36 0^{\circ} n, x = 198.42305 \dots^{\circ} + 36 0^{\circ} n, x = 45.94805 \dots^{\circ} + 36 0^{\circ} n, x = 134.05194 \dots^{\circ} + 36 0^{\circ} n

Pasos de solución

sin^{3} (x) - 2 sin (x) = 4 cos^{2} (x) - 3

Restar

4 cos^{2} (x) - 3

de ambos lados

sin^{3} (x) - 2 sin (x) - 4 cos^{2} (x) + 3 = 0

Re-escribir usando identidades trigonométricas

3 + sin^{3} (x) - 2 sin (x) - 4 cos^{2} (x)

Utilizar la identidad pitagórica:

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

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

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

Simplificar

3 + sin^{3} (x) - 2 sin (x) - 4 (1 - sin^{2} (x)) : sin^{3} (x) + 4 sin^{2} (x) - 2 sin (x) - 1

3 + sin^{3} (x) - 2 sin (x) - 4 (1 - sin^{2} (x))

Expandir

- 4 (1 - sin^{2} (x)) : - 4 + 4 sin^{2} (x)

- 4 (1 - sin^{2} (x))

Poner los parentesis utilizando:

a (b - c) = ab - a c

a = - 4, b = 1, c = sin^{2} (x)

= - 4 \cdot 1 - (- 4) sin^{2} (x)

Aplicar las reglas de los signos

- (- a) = a

= - 4 \cdot 1 + 4 sin^{2} (x)

Multiplicar los numeros:

4 \cdot 1 = 4

= - 4 + 4 sin^{2} (x)

= 3 + sin^{3} (x) - 2 sin (x) - 4 + 4 sin^{2} (x)

Simplificar

3 + sin^{3} (x) - 2 sin (x) - 4 + 4 sin^{2} (x) : sin^{3} (x) + 4 sin^{2} (x) - 2 sin (x) - 1

3 + sin^{3} (x) - 2 sin (x) - 4 + 4 sin^{2} (x)

Agrupar términos semejantes

= sin^{3} (x) - 2 sin (x) + 4 sin^{2} (x) + 3 - 4

Sumar/restar lo siguiente:

3 - 4 = - 1

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

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

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

- 1 + sin^{3} (x) - 2 sin (x) + 4 sin^{2} (x) = 0

Usando el método de sustitución

- 1 + sin^{3} (x) - 2 sin (x) + 4 sin^{2} (x) = 0

Sea:

sin (x) = u

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

- 1 + u^{3} - 2 u + 4 u^{2} = 0 : u \approx - 0.31603 \dots, u \approx 0.71870 \dots, u \approx - 4.40267 \dots

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

Escribir en la forma binómica

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

u^{3} + 4 u^{2} - 2 u - 1 = 0

Encontrar una solución para

u^{3} + 4 u^{2} - 2 u - 1 = 0

utilizando el método de Newton-Raphson:

u \approx - 0.31603 \dots

u^{3} + 4 u^{2} - 2 u - 1 = 0

Definición del método de Newton-Raphson

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

Hallar

f^{^{'}} (u) : 3 u^{2} + 8 u - 2

\frac{d}{d u} (u^{3} + 4 u^{2} - 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} (4 u^{2}) - \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} (4 u^{2}) = 8 u

\frac{d}{d u} (4 u^{2})

Sacar la constante:

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

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

Aplicar la regla de la potencia:

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

= 4 \cdot 2 u^{2 - 1}

Simplificar

= 8 u

\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} + 8 u - 2 - 0

Simplificar

= 3 u^{2} + 8 u - 2

Sea

u_{0} = 0

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = - 0.5 : Δ u_{1} = 0.5

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

f^{^{'}} (u_{0}) = 3 \cdot 0^{2} + 8 \cdot 0 - 2 = - 2

u_{1} = - 0.5

Δ u_{1} = ∣ - 0.5 - 0 ∣ = 0.5

Δ u_{1} = 0.5

u_{2} = - 0.33333 \dots : Δ u_{2} = 0.16666 \dots

f (u_{1}) = (- 0.5)^{3} + 4 (- 0.5)^{2} - 2 (- 0.5) - 1 = 0.875

f^{^{'}} (u_{1}) = 3 (- 0.5)^{2} + 8 (- 0.5) - 2 = - 5.25

u_{2} = - 0.33333 \dots

Δ u_{2} = ∣ - 0.33333 \dots - (- 0.5) ∣ = 0.16666 \dots

Δ u_{2} = 0.16666 \dots

u_{3} = - 0.31623 \dots : Δ u_{3} = 0.01709 \dots

f (u_{2}) = (- 0.33333 \dots)^{3} + 4 (- 0.33333 \dots)^{2} - 2 (- 0.33333 \dots) - 1 = 0.07407 \dots

f^{^{'}} (u_{2}) = 3 (- 0.33333 \dots)^{2} + 8 (- 0.33333 \dots) - 2 = - 4.33333 \dots

u_{3} = - 0.31623 \dots

Δ u_{3} = ∣ - 0.31623 \dots - (- 0.33333 \dots) ∣ = 0.01709 \dots

Δ u_{3} = 0.01709 \dots

u_{4} = - 0.31603 \dots : Δ u_{4} = 0.00020 \dots

f (u_{3}) = (- 0.31623 \dots)^{3} + 4 (- 0.31623 \dots)^{2} - 2 (- 0.31623 \dots) - 1 = 0.00088 \dots

f^{^{'}} (u_{3}) = 3 (- 0.31623 \dots)^{2} + 8 (- 0.31623 \dots) - 2 = - 4.22989 \dots

u_{4} = - 0.31603 \dots

Δ u_{4} = ∣ - 0.31603 \dots - (- 0.31623 \dots) ∣ = 0.00020 \dots

Δ u_{4} = 0.00020 \dots

u_{5} = - 0.31603 \dots : Δ u_{5} = 3.13479 E - 8

f (u_{4}) = (- 0.31603 \dots)^{3} + 4 (- 0.31603 \dots)^{2} - 2 (- 0.31603 \dots) - 1 = 1.32558 E - 7

f^{^{'}} (u_{4}) = 3 (- 0.31603 \dots)^{2} + 8 (- 0.31603 \dots) - 2 = - 4.22862 \dots

u_{5} = - 0.31603 \dots

Δ u_{5} = ∣ - 0.31603 \dots - (- 0.31603 \dots) ∣ = 3.13479 E - 8

Δ u_{5} = 3.13479 E - 8

u \approx - 0.31603 \dots

Aplicar la división larga

Eq u a t i o n 0 : \frac{u ^{3} + 4 u ^{2} - 2 u - 1}{u + 0.31603 \dots} = u^{2} + 3.68396 \dots u - 3.16424 \dots

u^{2} + 3.68396 \dots u - 3.16424 \dots \approx 0

Encontrar una solución para

u^{2} + 3.68396 \dots u - 3.16424 \dots = 0

utilizando el método de Newton-Raphson:

u \approx 0.71870 \dots

u^{2} + 3.68396 \dots u - 3.16424 \dots = 0

Definición del método de Newton-Raphson

f (u) = u^{2} + 3.68396 \dots u - 3.16424 \dots

Hallar

f^{^{'}} (u) : 2 u + 3.68396 \dots

\frac{d}{d u} (u^{2} + 3.68396 \dots u - 3.16424 \dots)

Aplicar la regla de la suma/diferencia:

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

= \frac{d}{d u} (u^{2}) + \frac{d}{d u} (3.68396 \dots u) - \frac{d}{d u} (3.16424 \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} (3.68396 \dots u) = 3.68396 \dots

\frac{d}{d u} (3.68396 \dots u)

Sacar la constante:

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

= 3.68396 \dots \frac{d u}{d u}

Aplicar la regla de derivación:

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

= 3.68396 \dots \cdot 1

Simplificar

= 3.68396 \dots

\frac{d}{d u} (3.16424 \dots) = 0

\frac{d}{d u} (3.16424 \dots)

Derivada de una constante:

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

= 0

= 2 u + 3.68396 \dots - 0

Simplificar

= 2 u + 3.68396 \dots

Sea

u_{0} = 1

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = 0.73263 \dots : Δ u_{1} = 0.26736 \dots

f (u_{0}) = 1^{2} + 3.68396 \dots \cdot 1 - 3.16424 \dots = 1.51972 \dots

f^{^{'}} (u_{0}) = 2 \cdot 1 + 3.68396 \dots = 5.68396 \dots

u_{1} = 0.73263 \dots

Δ u_{1} = ∣ 0.73263 \dots - 1 ∣ = 0.26736 \dots

Δ u_{1} = 0.26736 \dots

u_{2} = 0.71874 \dots : Δ u_{2} = 0.01388 \dots

f (u_{1}) = 0.73263 \dots^{2} + 3.68396 \dots \cdot 0.73263 \dots - 3.16424 \dots = 0.07148 \dots

f^{^{'}} (u_{1}) = 2 \cdot 0.73263 \dots + 3.68396 \dots = 5.14922 \dots

u_{2} = 0.71874 \dots

Δ u_{2} = ∣ 0.71874 \dots - 0.73263 \dots ∣ = 0.01388 \dots

Δ u_{2} = 0.01388 \dots

u_{3} = 0.71870 \dots : Δ u_{3} = 0.00003 \dots

f (u_{2}) = 0.71874 \dots^{2} + 3.68396 \dots \cdot 0.71874 \dots - 3.16424 \dots = 0.00019 \dots

f^{^{'}} (u_{2}) = 2 \cdot 0.71874 \dots + 3.68396 \dots = 5.12146 \dots

u_{3} = 0.71870 \dots

Δ u_{3} = ∣ 0.71870 \dots - 0.71874 \dots ∣ = 0.00003 \dots

Δ u_{3} = 0.00003 \dots

u_{4} = 0.71870 \dots : Δ u_{4} = 2.76537 E - 10

f (u_{3}) = 0.71870 \dots^{2} + 3.68396 \dots \cdot 0.71870 \dots - 3.16424 \dots = 1.41625 E - 9

f^{^{'}} (u_{3}) = 2 \cdot 0.71870 \dots + 3.68396 \dots = 5.12138 \dots

u_{4} = 0.71870 \dots

Δ u_{4} = ∣ 0.71870 \dots - 0.71870 \dots ∣ = 2.76537 E - 10

Δ u_{4} = 2.76537 E - 10

u \approx 0.71870 \dots

Aplicar la división larga

Eq u a t i o n 0 : \frac{u ^{2} + 3.68396 \dots u - 3.16424 \dots}{u - 0.71870 \dots} = u + 4.40267 \dots

u + 4.40267 \dots \approx 0

u \approx - 4.40267 \dots

Las soluciones son

u \approx - 0.31603 \dots, u \approx 0.71870 \dots, u \approx - 4.40267 \dots

Sustituir en la ecuación

u = sin (x)

sin (x) \approx - 0.31603 \dots, sin (x) \approx 0.71870 \dots, sin (x) \approx - 4.40267 \dots

sin (x) \approx - 0.31603 \dots, sin (x) \approx 0.71870 \dots, sin (x) \approx - 4.40267 \dots

sin (x) = - 0.31603 \dots : x = arcsin (- 0.31603 \dots) + 2 πn, x = π + arcsin (0.31603 \dots) + 2 πn

sin (x) = - 0.31603 \dots

Aplicar propiedades trigonométricas inversas

sin (x) = - 0.31603 \dots

Soluciones generales para

sin (x) = - 0.31603 \dots

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- 0.31603 \dots) + 2 πn, x = π + arcsin (0.31603 \dots) + 2 πn

x = arcsin (- 0.31603 \dots) + 2 πn, x = π + arcsin (0.31603 \dots) + 2 πn

sin (x) = 0.71870 \dots : x = arcsin (0.71870 \dots) + 2 πn, x = π - arcsin (0.71870 \dots) + 2 πn

sin (x) = 0.71870 \dots

Aplicar propiedades trigonométricas inversas

sin (x) = 0.71870 \dots

Soluciones generales para

sin (x) = 0.71870 \dots

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (0.71870 \dots) + 2 πn, x = π - arcsin (0.71870 \dots) + 2 πn

x = arcsin (0.71870 \dots) + 2 πn, x = π - arcsin (0.71870 \dots) + 2 πn

sin (x) = - 4.40267 \dots :

Sin solución

sin (x) = - 4.40267 \dots

- 1 \leq sin (x) \leq 1

Sin soluci \overset{o}{ˊ} n

Combinar toda las soluciones

x = arcsin (- 0.31603 \dots) + 2 πn, x = π + arcsin (0.31603 \dots) + 2 πn, x = arcsin (0.71870 \dots) + 2 πn, x = π - arcsin (0.71870 \dots) + 2 πn

Mostrar soluciones en forma decimal

x = - 0.32154 \dots + 2 πn, x = π + 0.32154 \dots + 2 πn, x = 0.80194 \dots + 2 πn, x = π - 0.80194 \dots + 2 πn

Gráfica

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