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

Solución

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

Solución

x = 0.85546 \dots + 2 πn, x = π - 0.85546 \dots + 2 πn

Grados

x = 49.01468 \dots^{\circ} + 36 0^{\circ} n, x = 130.98531 \dots^{\circ} + 36 0^{\circ} n

Pasos de solución

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

Restar

cos^{2} (x)

de ambos lados

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

Re-escribir usando identidades trigonométricas

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

Utilizar la identidad pitagórica:

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

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

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

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

Usando el método de sustitución

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

Sea:

sin (x) = u

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

- 1 + u^{2} + u^{3} = 0 : u \approx 0.75487 \dots

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

Escribir en la forma binómica

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

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

Encontrar una solución para

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

utilizando el método de Newton-Raphson:

u \approx 0.75487 \dots

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

Definición del método de Newton-Raphson

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

Hallar

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

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

Aplicar la regla de la suma/diferencia:

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

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

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

Derivada de una constante:

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

= 0

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

Simplificar

= 3 u^{2} + 2 u

Sea

u_{0} = 1

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = 0.8 : Δ u_{1} = 0.2

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

f^{^{'}} (u_{0}) = 3 \cdot 1^{2} + 2 \cdot 1 = 5

u_{1} = 0.8

Δ u_{1} = ∣ 0.8 - 1 ∣ = 0.2

Δ u_{1} = 0.2

u_{2} = 0.75681 \dots : Δ u_{2} = 0.04318 \dots

f (u_{1}) = 0. 8^{3} + 0. 8^{2} - 1 = 0.152

f^{^{'}} (u_{1}) = 3 \cdot 0. 8^{2} + 2 \cdot 0.8 = 3.52

u_{2} = 0.75681 \dots

Δ u_{2} = ∣ 0.75681 \dots - 0.8 ∣ = 0.04318 \dots

Δ u_{2} = 0.04318 \dots

u_{3} = 0.75488 \dots : Δ u_{3} = 0.00193 \dots

f (u_{2}) = 0.75681 \dots^{3} + 0.75681 \dots^{2} - 1 = 0.00625 \dots

f^{^{'}} (u_{2}) = 3 \cdot 0.75681 \dots^{2} + 2 \cdot 0.75681 \dots = 3.23195 \dots

u_{3} = 0.75488 \dots

Δ u_{3} = ∣ 0.75488 \dots - 0.75681 \dots ∣ = 0.00193 \dots

Δ u_{3} = 0.00193 \dots

u_{4} = 0.75487 \dots : Δ u_{4} = 3.80818 E - 6

f (u_{3}) = 0.75488 \dots^{3} + 0.75488 \dots^{2} - 1 = 0.00001 \dots

f^{^{'}} (u_{3}) = 3 \cdot 0.75488 \dots^{2} + 2 \cdot 0.75488 \dots = 3.21930 \dots

u_{4} = 0.75487 \dots

Δ u_{4} = ∣ 0.75487 \dots - 0.75488 \dots ∣ = 3.80818 E - 6

Δ u_{4} = 3.80818 E - 6

u_{5} = 0.75487 \dots : Δ u_{5} = 1.47065 E - 11

f (u_{4}) = 0.75487 \dots^{3} + 0.75487 \dots^{2} - 1 = 4.73444 E - 11

f^{^{'}} (u_{4}) = 3 \cdot 0.75487 \dots^{2} + 2 \cdot 0.75487 \dots = 3.21927 \dots

u_{5} = 0.75487 \dots

Δ u_{5} = ∣ 0.75487 \dots - 0.75487 \dots ∣ = 1.47065 E - 11

Δ u_{5} = 1.47065 E - 11

u \approx 0.75487 \dots

Aplicar la división larga

Eq u a t i o n 0 : \frac{u ^{3} + u ^{2} - 1}{u - 0.75487 \dots} = u^{2} + 1.75487 \dots u + 1.32471 \dots

u^{2} + 1.75487 \dots u + 1.32471 \dots \approx 0

Encontrar una solución para

u^{2} + 1.75487 \dots u + 1.32471 \dots = 0

utilizando el método de Newton-Raphson:

Sin solución para

u \in R

u^{2} + 1.75487 \dots u + 1.32471 \dots = 0

Definición del método de Newton-Raphson

f (u) = u^{2} + 1.75487 \dots u + 1.32471 \dots

Hallar

f^{^{'}} (u) : 2 u + 1.75487 \dots

\frac{d}{d u} (u^{2} + 1.75487 \dots u + 1.32471 \dots)

Aplicar la regla de la suma/diferencia:

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

= \frac{d}{d u} (u^{2}) + \frac{d}{d u} (1.75487 \dots u) + \frac{d}{d u} (1.32471 \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} (1.75487 \dots u) = 1.75487 \dots

\frac{d}{d u} (1.75487 \dots u)

Sacar la constante:

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

= 1.75487 \dots \frac{d u}{d u}

Aplicar la regla de derivación:

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

= 1.75487 \dots \cdot 1

Simplificar

= 1.75487 \dots

\frac{d}{d u} (1.32471 \dots) = 0

\frac{d}{d u} (1.32471 \dots)

Derivada de una constante:

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

= 0

= 2 u + 1.75487 \dots + 0

Simplificar

= 2 u + 1.75487 \dots

Sea

u_{0} = - 1

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = 1.32471 \dots : Δ u_{1} = 2.32471 \dots

f (u_{0}) = (- 1)^{2} + 1.75487 \dots (- 1) + 1.32471 \dots = 0.56984 \dots

f^{^{'}} (u_{0}) = 2 (- 1) + 1.75487 \dots = - 0.24512 \dots

u_{1} = 1.32471 \dots

Δ u_{1} = ∣ 1.32471 \dots - (- 1) ∣ = 2.32471 \dots

Δ u_{1} = 2.32471 \dots

u_{2} = 0.09766 \dots : Δ u_{2} = 1.22705 \dots

f (u_{1}) = 1.32471 \dots^{2} + 1.75487 \dots \cdot 1.32471 \dots + 1.32471 \dots = 5.40431 \dots

f^{^{'}} (u_{1}) = 2 \cdot 1.32471 \dots + 1.75487 \dots = 4.40431 \dots

u_{2} = 0.09766 \dots

Δ u_{2} = ∣ 0.09766 \dots - 1.32471 \dots ∣ = 1.22705 \dots

Δ u_{2} = 1.22705 \dots

u_{3} = - 0.67437 \dots : Δ u_{3} = 0.77204 \dots

f (u_{2}) = 0.09766 \dots^{2} + 1.75487 \dots \cdot 0.09766 \dots + 1.32471 \dots = 1.50565 \dots

f^{^{'}} (u_{2}) = 2 \cdot 0.09766 \dots + 1.75487 \dots = 1.95021 \dots

u_{3} = - 0.67437 \dots

Δ u_{3} = ∣ - 0.67437 \dots - 0.09766 \dots ∣ = 0.77204 \dots

Δ u_{3} = 0.77204 \dots

u_{4} = - 2.14204 \dots : Δ u_{4} = 1.46766 \dots

f (u_{3}) = (- 0.67437 \dots)^{2} + 1.75487 \dots (- 0.67437 \dots) + 1.32471 \dots = 0.59605 \dots

f^{^{'}} (u_{3}) = 2 (- 0.67437 \dots) + 1.75487 \dots = 0.40612 \dots

u_{4} = - 2.14204 \dots

Δ u_{4} = ∣ - 2.14204 \dots - (- 0.67437 \dots) ∣ = 1.46766 \dots

Δ u_{4} = 1.46766 \dots

u_{5} = - 1.29037 \dots : Δ u_{5} = 0.85166 \dots

f (u_{4}) = (- 2.14204 \dots)^{2} + 1.75487 \dots (- 2.14204 \dots) + 1.32471 \dots = 2.15403 \dots

f^{^{'}} (u_{4}) = 2 (- 2.14204 \dots) + 1.75487 \dots = - 2.52920 \dots

u_{5} = - 1.29037 \dots

Δ u_{5} = ∣ - 1.29037 \dots - (- 2.14204 \dots) ∣ = 0.85166 \dots

Δ u_{5} = 0.85166 \dots

u_{6} = - 0.41210 \dots : Δ u_{6} = 0.87826 \dots

f (u_{5}) = (- 1.29037 \dots)^{2} + 1.75487 \dots (- 1.29037 \dots) + 1.32471 \dots = 0.72533 \dots

f^{^{'}} (u_{5}) = 2 (- 1.29037 \dots) + 1.75487 \dots = - 0.82587 \dots

u_{6} = - 0.41210 \dots

Δ u_{6} = ∣ - 0.41210 \dots - (- 1.29037 \dots) ∣ = 0.87826 \dots

Δ u_{6} = 0.87826 \dots

u_{7} = - 1.24093 \dots : Δ u_{7} = 0.82882 \dots

f (u_{6}) = (- 0.41210 \dots)^{2} + 1.75487 \dots (- 0.41210 \dots) + 1.32471 \dots = 0.77135 \dots

f^{^{'}} (u_{6}) = 2 (- 0.41210 \dots) + 1.75487 \dots = 0.93065 \dots

u_{7} = - 1.24093 \dots

Δ u_{7} = ∣ - 1.24093 \dots - (- 0.41210 \dots) ∣ = 0.82882 \dots

Δ u_{7} = 0.82882 \dots

u_{8} = - 0.29600 \dots : Δ u_{8} = 0.94492 \dots

f (u_{7}) = (- 1.24093 \dots)^{2} + 1.75487 \dots (- 1.24093 \dots) + 1.32471 \dots = 0.68694 \dots

f^{^{'}} (u_{7}) = 2 (- 1.24093 \dots) + 1.75487 \dots = - 0.72698 \dots

u_{8} = - 0.29600 \dots

Δ u_{8} = ∣ - 0.29600 \dots - (- 1.24093 \dots) ∣ = 0.94492 \dots

Δ u_{8} = 0.94492 \dots

u_{9} = - 1.06383 \dots : Δ u_{9} = 0.76782 \dots

f (u_{8}) = (- 0.29600 \dots)^{2} + 1.75487 \dots (- 0.29600 \dots) + 1.32471 \dots = 0.89288 \dots

f^{^{'}} (u_{8}) = 2 (- 0.29600 \dots) + 1.75487 \dots = 1.16286 \dots

u_{9} = - 1.06383 \dots

Δ u_{9} = ∣ - 1.06383 \dots - (- 0.29600 \dots) ∣ = 0.76782 \dots

Δ u_{9} = 0.76782 \dots

u_{10} = 0.51763 \dots : Δ u_{10} = 1.58147 \dots

f (u_{9}) = (- 1.06383 \dots)^{2} + 1.75487 \dots (- 1.06383 \dots) + 1.32471 \dots = 0.58956 \dots

f^{^{'}} (u_{9}) = 2 (- 1.06383 \dots) + 1.75487 \dots = - 0.37279 \dots

u_{10} = 0.51763 \dots

Δ u_{10} = ∣ 0.51763 \dots - (- 1.06383 \dots) ∣ = 1.58147 \dots

Δ u_{10} = 1.58147 \dots

No se puede encontrar solución

La solución es

u \approx 0.75487 \dots

Sustituir en la ecuación

u = sin (x)

sin (x) \approx 0.75487 \dots

sin (x) \approx 0.75487 \dots

sin (x) = 0.75487 \dots : x = arcsin (0.75487 \dots) + 2 πn, x = π - arcsin (0.75487 \dots) + 2 πn

sin (x) = 0.75487 \dots

Aplicar propiedades trigonométricas inversas

sin (x) = 0.75487 \dots

Soluciones generales para

sin (x) = 0.75487 \dots

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

x = arcsin (0.75487 \dots) + 2 πn, x = π - arcsin (0.75487 \dots) + 2 πn

x = arcsin (0.75487 \dots) + 2 πn, x = π - arcsin (0.75487 \dots) + 2 πn

Combinar toda las soluciones

x = arcsin (0.75487 \dots) + 2 πn, x = π - arcsin (0.75487 \dots) + 2 πn

Mostrar soluciones en forma decimal

x = 0.85546 \dots + 2 πn, x = π - 0.85546 \dots + 2 πn

Ejemplos populares

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tan (x) = - 135

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

6=50sin(x)-15cos(x),0<x< pi/2

6 = 50 sin (x) - 15 cos (x), 0 < x < \frac{π}{2}