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sin^3(x)=cos(x)

Solución

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

Solución

x = 0.97202 \dots + 2 πn, x = π + 0.97202 \dots + 2 πn

Grados

x = 55.69319 \dots^{\circ} + 36 0^{\circ} n, x = 235.69319 \dots^{\circ} + 36 0^{\circ} n

Pasos de solución

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

Elevar al cuadrado ambos lados

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

Restar

cos^{2} (x)

de ambos lados

sin^{6} (x) - cos^{2} (x) = 0

Re-escribir usando identidades trigonométricas

- cos^{2} (x) + sin^{6} (x)

Utilizar la identidad pitagórica:

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

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

= - (1 - sin^{2} (x)) + sin^{6} (x)

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

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

Poner los parentesis

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

Aplicar las reglas de los signos

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

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

= - 1 + sin^{2} (x) + sin^{6} (x)

- 1 + sin^{2} (x) + sin^{6} (x) = 0

Usando el método de sustitución

- 1 + sin^{2} (x) + sin^{6} (x) = 0

Sea:

sin (x) = u

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

- 1 + u^{2} + u^{6} = 0 : u = 0.68232 \dots, u = - 0.68232 \dots

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

Escribir en la forma binómica

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

u^{6} + u^{2} - 1 = 0

Re-escribir la ecuación con

v = u^{2}

v^{3} = u^{6}

v^{3} + v - 1 = 0

Resolver

v^{3} + v - 1 = 0 : v \approx 0.68232 \dots

v^{3} + v - 1 = 0

Encontrar una solución para

v^{3} + v - 1 = 0

utilizando el método de Newton-Raphson:

v \approx 0.68232 \dots

v^{3} + v - 1 = 0

Definición del método de Newton-Raphson

f (v) = v^{3} + v - 1

Hallar

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

\frac{d}{d v} (v^{3} + v - 1)

Aplicar la regla de la suma/diferencia:

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

= \frac{d}{d v} (v^{3}) + \frac{d v}{d v} - \frac{d}{d v} (1)

\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 v}{d v} = 1

\frac{d v}{d v}

Aplicar la regla de derivación:

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

= 1

\frac{d}{d v} (1) = 0

\frac{d}{d v} (1)

Derivada de una constante:

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

= 0

= 3 v^{2} + 1 - 0

Simplificar

= 3 v^{2} + 1

Sea

v_{0} = 1

Calcular

v_{n + 1}

hasta que

Δ v_{n + 1} < 0.000001

v_{1} = 0.75 : Δ v_{1} = 0.25

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

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

v_{1} = 0.75

Δ v_{1} = ∣ 0.75 - 1 ∣ = 0.25

Δ v_{1} = 0.25

v_{2} = 0.68604 \dots : Δ v_{2} = 0.06395 \dots

f (v_{1}) = 0.7 5^{3} + 0.75 - 1 = 0.171875

f^{^{'}} (v_{1}) = 3 \cdot 0.7 5^{2} + 1 = 2.6875

v_{2} = 0.68604 \dots

Δ v_{2} = ∣ 0.68604 \dots - 0.75 ∣ = 0.06395 \dots

Δ v_{2} = 0.06395 \dots

v_{3} = 0.68233 \dots : Δ v_{3} = 0.00370 \dots

f (v_{2}) = 0.68604 \dots^{3} + 0.68604 \dots - 1 = 0.00894 \dots

f^{^{'}} (v_{2}) = 3 \cdot 0.68604 \dots^{2} + 1 = 2.41197 \dots

v_{3} = 0.68233 \dots

Δ v_{3} = ∣ 0.68233 \dots - 0.68604 \dots ∣ = 0.00370 \dots

Δ v_{3} = 0.00370 \dots

v_{4} = 0.68232 \dots : Δ v_{4} = 0.00001 \dots

f (v_{3}) = 0.68233 \dots^{3} + 0.68233 \dots - 1 = 0.00002 \dots

f^{^{'}} (v_{3}) = 3 \cdot 0.68233 \dots^{2} + 1 = 2.39676 \dots

v_{4} = 0.68232 \dots

Δ v_{4} = ∣ 0.68232 \dots - 0.68233 \dots ∣ = 0.00001 \dots

Δ v_{4} = 0.00001 \dots

v_{5} = 0.68232 \dots : Δ v_{5} = 1.18493 E - 10

f (v_{4}) = 0.68232 \dots^{3} + 0.68232 \dots - 1 = 2.83995 E - 10

f^{^{'}} (v_{4}) = 3 \cdot 0.68232 \dots^{2} + 1 = 2.39671 \dots

v_{5} = 0.68232 \dots

Δ v_{5} = ∣ 0.68232 \dots - 0.68232 \dots ∣ = 1.18493 E - 10

Δ v_{5} = 1.18493 E - 10

v \approx 0.68232 \dots

Aplicar la división larga

Eq u a t i o n 0 : \frac{v ^{3} + v - 1}{v - 0.68232 \dots} = v^{2} + 0.68232 \dots v + 1.46557 \dots

v^{2} + 0.68232 \dots v + 1.46557 \dots \approx 0

Encontrar una solución para

v^{2} + 0.68232 \dots v + 1.46557 \dots = 0

utilizando el método de Newton-Raphson:

Sin solución para

v \in R

v^{2} + 0.68232 \dots v + 1.46557 \dots = 0

Definición del método de Newton-Raphson

f (v) = v^{2} + 0.68232 \dots v + 1.46557 \dots

Hallar

f^{^{'}} (v) : 2 v + 0.68232 \dots

\frac{d}{d v} (v^{2} + 0.68232 \dots v + 1.46557 \dots)

Aplicar la regla de la suma/diferencia:

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

= \frac{d}{d v} (v^{2}) + \frac{d}{d v} (0.68232 \dots v) + \frac{d}{d v} (1.46557 \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} (0.68232 \dots v) = 0.68232 \dots

\frac{d}{d v} (0.68232 \dots v)

Sacar la constante:

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

= 0.68232 \dots \frac{d v}{d v}

Aplicar la regla de derivación:

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

= 0.68232 \dots \cdot 1

Simplificar

= 0.68232 \dots

\frac{d}{d v} (1.46557 \dots) = 0

\frac{d}{d v} (1.46557 \dots)

Derivada de una constante:

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

= 0

= 2 v + 0.68232 \dots + 0

Simplificar

= 2 v + 0.68232 \dots

Sea

v_{0} = - 2

Calcular

v_{n + 1}

hasta que

Δ v_{n + 1} < 0.000001

v_{1} = - 0.76391 \dots : Δ v_{1} = 1.23608 \dots

f (v_{0}) = (- 2)^{2} + 0.68232 \dots (- 2) + 1.46557 \dots = 4.10091 \dots

f^{^{'}} (v_{0}) = 2 (- 2) + 0.68232 \dots = - 3.31767 \dots

v_{1} = - 0.76391 \dots

Δ v_{1} = ∣ - 0.76391 \dots - (- 2) ∣ = 1.23608 \dots

Δ v_{1} = 1.23608 \dots

v_{2} = 1.04316 \dots : Δ v_{2} = 1.80707 \dots

f (v_{1}) = (- 0.76391 \dots)^{2} + 0.68232 \dots (- 0.76391 \dots) + 1.46557 \dots = 1.52789 \dots

f^{^{'}} (v_{1}) = 2 (- 0.76391 \dots) + 0.68232 \dots = - 0.84550 \dots

v_{2} = 1.04316 \dots

Δ v_{2} = ∣ 1.04316 \dots - (- 0.76391 \dots) ∣ = 1.80707 \dots

Δ v_{2} = 1.80707 \dots

v_{3} = - 0.13630 \dots : Δ v_{3} = 1.17946 \dots

f (v_{2}) = 1.04316 \dots^{2} + 0.68232 \dots \cdot 1.04316 \dots + 1.46557 \dots = 3.26553 \dots

f^{^{'}} (v_{2}) = 2 \cdot 1.04316 \dots + 0.68232 \dots = 2.76865 \dots

v_{3} = - 0.13630 \dots

Δ v_{3} = ∣ - 0.13630 \dots - 1.04316 \dots ∣ = 1.17946 \dots

Δ v_{3} = 1.17946 \dots

v_{4} = - 3.53171 \dots : Δ v_{4} = 3.39540 \dots

f (v_{3}) = (- 0.13630 \dots)^{2} + 0.68232 \dots (- 0.13630 \dots) + 1.46557 \dots = 1.39114 \dots

f^{^{'}} (v_{3}) = 2 (- 0.13630 \dots) + 0.68232 \dots = 0.40971 \dots

v_{4} = - 3.53171 \dots

Δ v_{4} = ∣ - 3.53171 \dots - (- 0.13630 \dots) ∣ = 3.39540 \dots

Δ v_{4} = 3.39540 \dots

v_{5} = - 1.72500 \dots : Δ v_{5} = 1.80670 \dots

f (v_{4}) = (- 3.53171 \dots)^{2} + 0.68232 \dots (- 3.53171 \dots) + 1.46557 \dots = 11.52876 \dots

f^{^{'}} (v_{4}) = 2 (- 3.53171 \dots) + 0.68232 \dots = - 6.38109 \dots

v_{5} = - 1.72500 \dots

Δ v_{5} = ∣ - 1.72500 \dots - (- 3.53171 \dots) ∣ = 1.80670 \dots

Δ v_{5} = 1.80670 \dots

v_{6} = - 0.54560 \dots : Δ v_{6} = 1.17939 \dots

f (v_{5}) = (- 1.72500 \dots)^{2} + 0.68232 \dots (- 1.72500 \dots) + 1.46557 \dots = 3.26419 \dots

f^{^{'}} (v_{5}) = 2 (- 1.72500 \dots) + 0.68232 \dots = - 2.76767 \dots

v_{6} = - 0.54560 \dots

Δ v_{6} = ∣ - 0.54560 \dots - (- 1.72500 \dots) ∣ = 1.17939 \dots

Δ v_{6} = 1.17939 \dots

v_{7} = 2.85625 \dots : Δ v_{7} = 3.40185 \dots

f (v_{6}) = (- 0.54560 \dots)^{2} + 0.68232 \dots (- 0.54560 \dots) + 1.46557 \dots = 1.39097 \dots

f^{^{'}} (v_{6}) = 2 (- 0.54560 \dots) + 0.68232 \dots = - 0.40888 \dots

v_{7} = 2.85625 \dots

Δ v_{7} = ∣ 2.85625 \dots - (- 0.54560 \dots) ∣ = 3.40185 \dots

Δ v_{7} = 3.40185 \dots

v_{8} = 1.04656 \dots : Δ v_{8} = 1.80968 \dots

f (v_{7}) = 2.85625 \dots^{2} + 0.68232 \dots \cdot 2.85625 \dots + 1.46557 \dots = 11.57264 \dots

f^{^{'}} (v_{7}) = 2 \cdot 2.85625 \dots + 0.68232 \dots = 6.39483 \dots

v_{8} = 1.04656 \dots

Δ v_{8} = ∣ 1.04656 \dots - 2.85625 \dots ∣ = 1.80968 \dots

Δ v_{8} = 1.80968 \dots

v_{9} = - 0.13340 \dots : Δ v_{9} = 1.17997 \dots

f (v_{8}) = 1.04656 \dots^{2} + 0.68232 \dots \cdot 1.04656 \dots + 1.46557 \dots = 3.27496 \dots

f^{^{'}} (v_{8}) = 2 \cdot 1.04656 \dots + 0.68232 \dots = 2.77545 \dots

v_{9} = - 0.13340 \dots

Δ v_{9} = ∣ - 0.13340 \dots - 1.04656 \dots ∣ = 1.17997 \dots

Δ v_{9} = 1.17997 \dots

v_{10} = - 3.48434 \dots : Δ v_{10} = 3.35093 \dots

f (v_{9}) = (- 0.13340 \dots)^{2} + 0.68232 \dots (- 0.13340 \dots) + 1.46557 \dots = 1.39234 \dots

f^{^{'}} (v_{9}) = 2 (- 0.13340 \dots) + 0.68232 \dots = 0.41550 \dots

v_{10} = - 3.48434 \dots

Δ v_{10} = ∣ - 3.48434 \dots - (- 0.13340 \dots) ∣ = 3.35093 \dots

Δ v_{10} = 3.35093 \dots

No se puede encontrar solución

La solución es

v \approx 0.68232 \dots

v \approx 0.68232 \dots

Sustituir hacia atrás la

v = u^{2},

resolver para

u

Resolver

u^{2} = 0.68232 \dots : u = 0.68232 \dots, u = - 0.68232 \dots

u^{2} = 0.68232 \dots

Para

x^{2} = f (a)

las soluciones son

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

u = 0.68232 \dots, u = - 0.68232 \dots

Las soluciones son

u = 0.68232 \dots, u = - 0.68232 \dots

Sustituir en la ecuación

u = sin (x)

sin (x) = 0.68232 \dots, sin (x) = - 0.68232 \dots

sin (x) = 0.68232 \dots, sin (x) = - 0.68232 \dots

sin (x) = 0.68232 \dots : x = arcsin (0.68232 \dots) + 2 πn, x = π - arcsin (0.68232 \dots) + 2 πn

sin (x) = 0.68232 \dots

Aplicar propiedades trigonométricas inversas

sin (x) = 0.68232 \dots

Soluciones generales para

sin (x) = 0.68232 \dots

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

x = arcsin (0.68232 \dots) + 2 πn, x = π - arcsin (0.68232 \dots) + 2 πn

x = arcsin (0.68232 \dots) + 2 πn, x = π - arcsin (0.68232 \dots) + 2 πn

sin (x) = - 0.68232 \dots : x = arcsin (- 0.68232 \dots) + 2 πn, x = π + arcsin (0.68232 \dots) + 2 πn

sin (x) = - 0.68232 \dots

Aplicar propiedades trigonométricas inversas

sin (x) = - 0.68232 \dots

Soluciones generales para

sin (x) = - 0.68232 \dots

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

x = arcsin (- 0.68232 \dots) + 2 πn, x = π + arcsin (0.68232 \dots) + 2 πn

x = arcsin (- 0.68232 \dots) + 2 πn, x = π + arcsin (0.68232 \dots) + 2 πn

Combinar toda las soluciones

x = arcsin (0.68232 \dots) + 2 πn, x = π - arcsin (0.68232 \dots) + 2 πn, x = arcsin (- 0.68232 \dots) + 2 πn, x = π + arcsin (0.68232 \dots) + 2 πn

Verificar las soluciones sustituyendo en la ecuación original

Verificar las soluciones sustituyéndolas en

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

Quitar las que no concuerden con la ecuación.

Verificar la solución

arcsin (0.68232 \dots) + 2 πn :

Verdadero

arcsin (0.68232 \dots) + 2 πn

Sustituir

n = 1

arcsin (0.68232 \dots) + 2 π 1

Multiplicar

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

por

x = arcsin (0.68232 \dots) + 2 π 1

sin^{3} (arcsin (0.68232 \dots) + 2 π 1) = cos (arcsin (0.68232 \dots) + 2 π 1)

Simplificar

0.56362 \dots = 0.56362 \dots

\Rightarrow Verdadero

Verificar la solución

π - arcsin (0.68232 \dots) + 2 πn :

Falso

π - arcsin (0.68232 \dots) + 2 πn

Sustituir

n = 1

π - arcsin (0.68232 \dots) + 2 π 1

Multiplicar

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

por

x = π - arcsin (0.68232 \dots) + 2 π 1

sin^{3} (π - arcsin (0.68232 \dots) + 2 π 1) = cos (π - arcsin (0.68232 \dots) + 2 π 1)

Simplificar

0.56362 \dots = - 0.56362 \dots

\Rightarrow Falso

Verificar la solución

arcsin (- 0.68232 \dots) + 2 πn :

Falso

arcsin (- 0.68232 \dots) + 2 πn

Sustituir

n = 1

arcsin (- 0.68232 \dots) + 2 π 1

Multiplicar

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

por

x = arcsin (- 0.68232 \dots) + 2 π 1

sin^{3} (arcsin (- 0.68232 \dots) + 2 π 1) = cos (arcsin (- 0.68232 \dots) + 2 π 1)

Simplificar

- 0.56362 \dots = 0.56362 \dots

\Rightarrow Falso

Verificar la solución

π + arcsin (0.68232 \dots) + 2 πn :

Verdadero

π + arcsin (0.68232 \dots) + 2 πn

Sustituir

n = 1

π + arcsin (0.68232 \dots) + 2 π 1

Multiplicar

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

por

x = π + arcsin (0.68232 \dots) + 2 π 1

sin^{3} (π + arcsin (0.68232 \dots) + 2 π 1) = cos (π + arcsin (0.68232 \dots) + 2 π 1)

Simplificar

- 0.56362 \dots = - 0.56362 \dots

\Rightarrow Verdadero

x = arcsin (0.68232 \dots) + 2 πn, x = π + arcsin (0.68232 \dots) + 2 πn

Mostrar soluciones en forma decimal

x = 0.97202 \dots + 2 πn, x = π + 0.97202 \dots + 2 πn

Ejemplos populares

tan((91pi)/(36))=cos(360-x)

tan (\frac{91 π}{36}) = cos (36 0^{\circ} - x)

2sin(3x-pi/6)+sqrt(3)=0

2 sin (3 x - \frac{π}{6}) + 3 = 0

12^2=10^2+12^2-2(10)(12)cos(b)

1 2^{2} = 1 0^{2} + 1 2^{2} - 2 (10) (12) cos (b)

solvefor x,sin(t)-csc(x)+1=0 resolver para

x, sin (t) - csc (x) + 1 = 0

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

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