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sin^3(x)+sin(x)=2cos^2(x)

Solución

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

Solución

x = 0.76928 \dots + 2 πn, x = π - 0.76928 \dots + 2 πn

Grados

x = 44.07670 \dots^{\circ} + 36 0^{\circ} n, x = 135.92329 \dots^{\circ} + 36 0^{\circ} n

Pasos de solución

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

Restar

2 cos^{2} (x)

de ambos lados

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

Re-escribir usando identidades trigonométricas

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

Utilizar la identidad pitagórica:

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

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

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

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

Usando el método de sustitución

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

Sea:

sin (x) = u

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

u + u^{3} - (1 - u^{2}) \cdot 2 = 0 : u \approx 0.69562 \dots

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

Desarrollar

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

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

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

Expandir

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

- 2 (1 - u^{2})

Poner los parentesis utilizando:

a (b - c) = ab - a c

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

= - 2 \cdot 1 - (- 2) u^{2}

Aplicar las reglas de los signos

- (- a) = a

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

Multiplicar los numeros:

2 \cdot 1 = 2

= - 2 + 2 u^{2}

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

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

Escribir en la forma binómica

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

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

Encontrar una solución para

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

utilizando el método de Newton-Raphson:

u \approx 0.69562 \dots

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

Definición del método de Newton-Raphson

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

Hallar

f^{^{'}} (u) : 3 u^{2} + 4 u + 1

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

Aplicar la regla de la suma/diferencia:

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

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

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

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

Sacar la constante:

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

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

Aplicar la regla de la potencia:

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

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

Simplificar

= 4 u

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

\frac{d u}{d u}

Aplicar la regla de derivación:

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

= 1

\frac{d}{d u} (2) = 0

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

Derivada de una constante:

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

= 0

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

Simplificar

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

Sea

u_{0} = 2

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = 1.23809 \dots : Δ u_{1} = 0.76190 \dots

f (u_{0}) = 2^{3} + 2 \cdot 2^{2} + 2 - 2 = 16

f^{^{'}} (u_{0}) = 3 \cdot 2^{2} + 4 \cdot 2 + 1 = 21

u_{1} = 1.23809 \dots

Δ u_{1} = ∣ 1.23809 \dots - 2 ∣ = 0.76190 \dots

Δ u_{1} = 0.76190 \dots

u_{2} = 0.83986 \dots : Δ u_{2} = 0.39822 \dots

f (u_{1}) = 1.23809 \dots^{3} + 2 \cdot 1.23809 \dots^{2} + 1.23809 \dots - 2 = 4.20170 \dots

f^{^{'}} (u_{1}) = 3 \cdot 1.23809 \dots^{2} + 4 \cdot 1.23809 \dots + 1 = 10.55102 \dots

u_{2} = 0.83986 \dots

Δ u_{2} = ∣ 0.83986 \dots - 1.23809 \dots ∣ = 0.39822 \dots

Δ u_{2} = 0.39822 \dots

u_{3} = 0.70967 \dots : Δ u_{3} = 0.13018 \dots

f (u_{2}) = 0.83986 \dots^{3} + 2 \cdot 0.83986 \dots^{2} + 0.83986 \dots - 2 = 0.84304 \dots

f^{^{'}} (u_{2}) = 3 \cdot 0.83986 \dots^{2} + 4 \cdot 0.83986 \dots + 1 = 6.47560 \dots

u_{3} = 0.70967 \dots

Δ u_{3} = ∣ 0.70967 \dots - 0.83986 \dots ∣ = 0.13018 \dots

Δ u_{3} = 0.13018 \dots

u_{4} = 0.69577 \dots : Δ u_{4} = 0.01390 \dots

f (u_{3}) = 0.70967 \dots^{3} + 2 \cdot 0.70967 \dots^{2} + 0.70967 \dots - 2 = 0.07439 \dots

f^{^{'}} (u_{3}) = 3 \cdot 0.70967 \dots^{2} + 4 \cdot 0.70967 \dots + 1 = 5.34965 \dots

u_{4} = 0.69577 \dots

Δ u_{4} = ∣ 0.69577 \dots - 0.70967 \dots ∣ = 0.01390 \dots

Δ u_{4} = 0.01390 \dots

u_{5} = 0.69562 \dots : Δ u_{5} = 0.00015 \dots

f (u_{4}) = 0.69577 \dots^{3} + 2 \cdot 0.69577 \dots^{2} + 0.69577 \dots - 2 = 0.00079 \dots

f^{^{'}} (u_{4}) = 3 \cdot 0.69577 \dots^{2} + 4 \cdot 0.69577 \dots + 1 = 5.23539 \dots

u_{5} = 0.69562 \dots

Δ u_{5} = ∣ 0.69562 \dots - 0.69577 \dots ∣ = 0.00015 \dots

Δ u_{5} = 0.00015 \dots

u_{6} = 0.69562 \dots : Δ u_{6} = 1.80445 E - 8

f (u_{5}) = 0.69562 \dots^{3} + 2 \cdot 0.69562 \dots^{2} + 0.69562 \dots - 2 = 9.44475 E - 8

f^{^{'}} (u_{5}) = 3 \cdot 0.69562 \dots^{2} + 4 \cdot 0.69562 \dots + 1 = 5.23414 \dots

u_{6} = 0.69562 \dots

Δ u_{6} = ∣ 0.69562 \dots - 0.69562 \dots ∣ = 1.80445 E - 8

Δ u_{6} = 1.80445 E - 8

u \approx 0.69562 \dots

Aplicar la división larga

Eq u a t i o n 0 : \frac{u ^{3} + 2 u ^{2} + u - 2}{u - 0.69562 \dots} = u^{2} + 2.69562 \dots u + 2.87512 \dots

u^{2} + 2.69562 \dots u + 2.87512 \dots \approx 0

Encontrar una solución para

u^{2} + 2.69562 \dots u + 2.87512 \dots = 0

utilizando el método de Newton-Raphson:

Sin solución para

u \in R

u^{2} + 2.69562 \dots u + 2.87512 \dots = 0

Definición del método de Newton-Raphson

f (u) = u^{2} + 2.69562 \dots u + 2.87512 \dots

Hallar

f^{^{'}} (u) : 2 u + 2.69562 \dots

\frac{d}{d u} (u^{2} + 2.69562 \dots u + 2.87512 \dots)

Aplicar la regla de la suma/diferencia:

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

= \frac{d}{d u} (u^{2}) + \frac{d}{d u} (2.69562 \dots u) + \frac{d}{d u} (2.87512 \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} (2.69562 \dots u) = 2.69562 \dots

\frac{d}{d u} (2.69562 \dots u)

Sacar la constante:

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

= 2.69562 \dots \frac{d u}{d u}

Aplicar la regla de derivación:

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

= 2.69562 \dots \cdot 1

Simplificar

= 2.69562 \dots

\frac{d}{d u} (2.87512 \dots) = 0

\frac{d}{d u} (2.87512 \dots)

Derivada de una constante:

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

= 0

= 2 u + 2.69562 \dots + 0

Simplificar

= 2 u + 2.69562 \dots

Sea

u_{0} = - 1

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = - 2.69562 \dots : Δ u_{1} = 1.69562 \dots

f (u_{0}) = (- 1)^{2} + 2.69562 \dots (- 1) + 2.87512 \dots = 1.17950 \dots

f^{^{'}} (u_{0}) = 2 (- 1) + 2.69562 \dots = 0.69562 \dots

u_{1} = - 2.69562 \dots

Δ u_{1} = ∣ - 2.69562 \dots - (- 1) ∣ = 1.69562 \dots

Δ u_{1} = 1.69562 \dots

u_{2} = - 1.62902 \dots : Δ u_{2} = 1.06659 \dots

f (u_{1}) = (- 2.69562 \dots)^{2} + 2.69562 \dots (- 2.69562 \dots) + 2.87512 \dots = 2.87512 \dots

f^{^{'}} (u_{1}) = 2 (- 2.69562 \dots) + 2.69562 \dots = - 2.69562 \dots

u_{2} = - 1.62902 \dots

Δ u_{2} = ∣ - 1.62902 \dots - (- 2.69562 \dots) ∣ = 1.06659 \dots

Δ u_{2} = 1.06659 \dots

u_{3} = 0.39364 \dots : Δ u_{3} = 2.02266 \dots

f (u_{2}) = (- 1.62902 \dots)^{2} + 2.69562 \dots (- 1.62902 \dots) + 2.87512 \dots = 1.13762 \dots

f^{^{'}} (u_{2}) = 2 (- 1.62902 \dots) + 2.69562 \dots = - 0.56243 \dots

u_{3} = 0.39364 \dots

Δ u_{3} = ∣ 0.39364 \dots - (- 1.62902 \dots) ∣ = 2.02266 \dots

Δ u_{3} = 2.02266 \dots

u_{4} = - 0.78100 \dots : Δ u_{4} = 1.17464 \dots

f (u_{3}) = 0.39364 \dots^{2} + 2.69562 \dots \cdot 0.39364 \dots + 2.87512 \dots = 4.09119 \dots

f^{^{'}} (u_{3}) = 2 \cdot 0.39364 \dots + 2.69562 \dots = 3.48290 \dots

u_{4} = - 0.78100 \dots

Δ u_{4} = ∣ - 0.78100 \dots - 0.39364 \dots ∣ = 1.17464 \dots

Δ u_{4} = 1.17464 \dots

u_{5} = - 1.99818 \dots : Δ u_{5} = 1.21718 \dots

f (u_{4}) = (- 0.78100 \dots)^{2} + 2.69562 \dots (- 0.78100 \dots) + 2.87512 \dots = 1.37980 \dots

f^{^{'}} (u_{4}) = 2 (- 0.78100 \dots) + 2.69562 \dots = 1.13360 \dots

u_{5} = - 1.99818 \dots

Δ u_{5} = ∣ - 1.99818 \dots - (- 0.78100 \dots) ∣ = 1.21718 \dots

Δ u_{5} = 1.21718 \dots

u_{6} = - 0.85921 \dots : Δ u_{6} = 1.13897 \dots

f (u_{5}) = (- 1.99818 \dots)^{2} + 2.69562 \dots (- 1.99818 \dots) + 2.87512 \dots = 1.48152 \dots

f^{^{'}} (u_{5}) = 2 (- 1.99818 \dots) + 2.69562 \dots = - 1.30075 \dots

u_{6} = - 0.85921 \dots

Δ u_{6} = ∣ - 0.85921 \dots - (- 1.99818 \dots) ∣ = 1.13897 \dots

Δ u_{6} = 1.13897 \dots

u_{7} = - 2.18675 \dots : Δ u_{7} = 1.32754 \dots

f (u_{6}) = (- 0.85921 \dots)^{2} + 2.69562 \dots (- 0.85921 \dots) + 2.87512 \dots = 1.29726 \dots

f^{^{'}} (u_{6}) = 2 (- 0.85921 \dots) + 2.69562 \dots = 0.97719 \dots

u_{7} = - 2.18675 \dots

Δ u_{7} = ∣ - 2.18675 \dots - (- 0.85921 \dots) ∣ = 1.32754 \dots

Δ u_{7} = 1.32754 \dots

u_{8} = - 1.13641 \dots : Δ u_{8} = 1.05034 \dots

f (u_{7}) = (- 2.18675 \dots)^{2} + 2.69562 \dots (- 2.18675 \dots) + 2.87512 \dots = 1.76236 \dots

f^{^{'}} (u_{7}) = 2 (- 2.18675 \dots) + 2.69562 \dots = - 1.67789 \dots

u_{8} = - 1.13641 \dots

Δ u_{8} = ∣ - 1.13641 \dots - (- 2.18675 \dots) ∣ = 1.05034 \dots

Δ u_{8} = 1.05034 \dots

u_{9} = - 3.74574 \dots : Δ u_{9} = 2.60933 \dots

f (u_{8}) = (- 1.13641 \dots)^{2} + 2.69562 \dots (- 1.13641 \dots) + 2.87512 \dots = 1.10322 \dots

f^{^{'}} (u_{8}) = 2 (- 1.13641 \dots) + 2.69562 \dots = 0.42279 \dots

u_{9} = - 3.74574 \dots

Δ u_{9} = ∣ - 3.74574 \dots - (- 1.13641 \dots) ∣ = 2.60933 \dots

Δ u_{9} = 2.60933 \dots

No se puede encontrar solución

La solución es

u \approx 0.69562 \dots

Sustituir en la ecuación

u = sin (x)

sin (x) \approx 0.69562 \dots

sin (x) \approx 0.69562 \dots

sin (x) = 0.69562 \dots : x = arcsin (0.69562 \dots) + 2 πn, x = π - arcsin (0.69562 \dots) + 2 πn

sin (x) = 0.69562 \dots

Aplicar propiedades trigonométricas inversas

sin (x) = 0.69562 \dots

Soluciones generales para

sin (x) = 0.69562 \dots

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

x = arcsin (0.69562 \dots) + 2 πn, x = π - arcsin (0.69562 \dots) + 2 πn

x = arcsin (0.69562 \dots) + 2 πn, x = π - arcsin (0.69562 \dots) + 2 πn

Combinar toda las soluciones

x = arcsin (0.69562 \dots) + 2 πn, x = π - arcsin (0.69562 \dots) + 2 πn

Mostrar soluciones en forma decimal

x = 0.76928 \dots + 2 πn, x = π - 0.76928 \dots + 2 πn

Ejemplos populares

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tan^2(x)cos(x)+cos^2(x)=1

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

sinh(x)= 3/2

sinh (x) = \frac{3}{2}

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a = 2 sin (3 0^{\circ} + x) - cos (x)