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cos(x)+cos^4(x)= 1/2

Solución

cos (x) + cos^{4} (x) = \frac{1}{2}

Solución

x = 1.09667 \dots + 2 πn, x = 2 π - 1.09667 \dots + 2 πn

Grados

x = 62.83512 \dots^{\circ} + 36 0^{\circ} n, x = 297.16487 \dots^{\circ} + 36 0^{\circ} n

Pasos de solución

cos (x) + cos^{4} (x) = \frac{1}{2}

Usando el método de sustitución

cos (x) + cos^{4} (x) = \frac{1}{2}

Sea:

cos (x) = u

u + u^{4} = \frac{1}{2}

u + u^{4} = \frac{1}{2} : u \approx 0.45655 \dots, u \approx - 1.12990 \dots

u + u^{4} = \frac{1}{2}

Multiplicar ambos lados por

2

u + u^{4} = \frac{1}{2}

Multiplicar ambos lados por

2

u \cdot 2 + u^{4} \cdot 2 = \frac{1}{2} \cdot 2

Simplificar

2 u + 2 u^{4} = 1

2 u + 2 u^{4} = 1

Mover

1

al lado izquierdo

2 u + 2 u^{4} = 1

Restar

1

de ambos lados

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

Simplificar

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

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

Escribir en la forma binómica

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

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

Encontrar una solución para

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

utilizando el método de Newton-Raphson:

u \approx 0.45655 \dots

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

Definición del método de Newton-Raphson

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

Hallar

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

\frac{d}{d u} (2 u^{4} + 2 u - 1)

Aplicar la regla de la suma/diferencia:

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

= \frac{d}{d u} (2 u^{4}) + \frac{d}{d u} (2 u) - \frac{d}{d u} (1)

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

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

Sacar la constante:

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

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

Aplicar la regla de la potencia:

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

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

Simplificar

= 8 u^{3}

\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

= 8 u^{3} + 2 - 0

Simplificar

= 8 u^{3} + 2

Sea

u_{0} = 1

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = 0.7 : Δ u_{1} = 0.3

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

f^{^{'}} (u_{0}) = 8 \cdot 1^{3} + 2 = 10

u_{1} = 0.7

Δ u_{1} = ∣ 0.7 - 1 ∣ = 0.3

Δ u_{1} = 0.3

u_{2} = 0.51446 \dots : Δ u_{2} = 0.18553 \dots

f (u_{1}) = 2 \cdot 0. 7^{4} + 2 \cdot 0.7 - 1 = 0.8802

f^{^{'}} (u_{1}) = 8 \cdot 0. 7^{3} + 2 = 4.744

u_{2} = 0.51446 \dots

Δ u_{2} = ∣ 0.51446 \dots - 0.7 ∣ = 0.18553 \dots

Δ u_{2} = 0.18553 \dots

u_{3} = 0.45974 \dots : Δ u_{3} = 0.05471 \dots

f (u_{2}) = 2 \cdot 0.51446 \dots^{4} + 2 \cdot 0.51446 \dots - 1 = 0.16902 \dots

f^{^{'}} (u_{2}) = 8 \cdot 0.51446 \dots^{3} + 2 = 3.08929 \dots

u_{3} = 0.45974 \dots

Δ u_{3} = ∣ 0.45974 \dots - 0.51446 \dots ∣ = 0.05471 \dots

Δ u_{3} = 0.05471 \dots

u_{4} = 0.45656 \dots : Δ u_{4} = 0.00318 \dots

f (u_{3}) = 2 \cdot 0.45974 \dots^{4} + 2 \cdot 0.45974 \dots - 1 = 0.00885 \dots

f^{^{'}} (u_{3}) = 8 \cdot 0.45974 \dots^{3} + 2 = 2.77741 \dots

u_{4} = 0.45656 \dots

Δ u_{4} = ∣ 0.45656 \dots - 0.45974 \dots ∣ = 0.00318 \dots

Δ u_{4} = 0.00318 \dots

u_{5} = 0.45655 \dots : Δ u_{5} = 9.28514 E - 6

f (u_{4}) = 2 \cdot 0.45656 \dots^{4} + 2 \cdot 0.45656 \dots - 1 = 0.00002 \dots

f^{^{'}} (u_{4}) = 8 \cdot 0.45656 \dots^{3} + 2 = 2.76135 \dots

u_{5} = 0.45655 \dots

Δ u_{5} = ∣ 0.45655 \dots - 0.45656 \dots ∣ = 9.28514 E - 6

Δ u_{5} = 9.28514 E - 6

u_{6} = 0.45655 \dots : Δ u_{6} = 7.80973 E - 11

f (u_{5}) = 2 \cdot 0.45655 \dots^{4} + 2 \cdot 0.45655 \dots - 1 = 2.15651 E - 10

f^{^{'}} (u_{5}) = 8 \cdot 0.45655 \dots^{3} + 2 = 2.76131 \dots

u_{6} = 0.45655 \dots

Δ u_{6} = ∣ 0.45655 \dots - 0.45655 \dots ∣ = 7.80973 E - 11

Δ u_{6} = 7.80973 E - 11

u \approx 0.45655 \dots

Aplicar la división larga

Eq u a t i o n 0 : \frac{2 u ^{4} + 2 u - 1}{u - 0.45655 \dots} = 2 u^{3} + 0.91310 \dots u^{2} + 0.41688 \dots u + 2.19032 \dots

2 u^{3} + 0.91310 \dots u^{2} + 0.41688 \dots u + 2.19032 \dots \approx 0

Encontrar una solución para

2 u^{3} + 0.91310 \dots u^{2} + 0.41688 \dots u + 2.19032 \dots = 0

utilizando el método de Newton-Raphson:

u \approx - 1.12990 \dots

2 u^{3} + 0.91310 \dots u^{2} + 0.41688 \dots u + 2.19032 \dots = 0

Definición del método de Newton-Raphson

f (u) = 2 u^{3} + 0.91310 \dots u^{2} + 0.41688 \dots u + 2.19032 \dots

Hallar

f^{^{'}} (u) : 6 u^{2} + 1.82621 \dots u + 0.41688 \dots

\frac{d}{d u} (2 u^{3} + 0.91310 \dots u^{2} + 0.41688 \dots u + 2.19032 \dots)

Aplicar la regla de la suma/diferencia:

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

= \frac{d}{d u} (2 u^{3}) + \frac{d}{d u} (0.91310 \dots u^{2}) + \frac{d}{d u} (0.41688 \dots u) + \frac{d}{d u} (2.19032 \dots)

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

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

Sacar la constante:

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

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

Aplicar la regla de la potencia:

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

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

Simplificar

= 6 u^{2}

\frac{d}{d u} (0.91310 \dots u^{2}) = 1.82621 \dots u

\frac{d}{d u} (0.91310 \dots u^{2})

Sacar la constante:

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

= 0.91310 \dots \frac{d}{d u} (u^{2})

Aplicar la regla de la potencia:

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

= 0.91310 \dots \cdot 2 u^{2 - 1}

Simplificar

= 1.82621 \dots u

\frac{d}{d u} (0.41688 \dots u) = 0.41688 \dots

\frac{d}{d u} (0.41688 \dots u)

Sacar la constante:

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

= 0.41688 \dots \frac{d u}{d u}

Aplicar la regla de derivación:

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

= 0.41688 \dots \cdot 1

Simplificar

= 0.41688 \dots

\frac{d}{d u} (2.19032 \dots) = 0

\frac{d}{d u} (2.19032 \dots)

Derivada de una constante:

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

= 0

= 6 u^{2} + 1.82621 \dots u + 0.41688 \dots + 0

Simplificar

= 6 u^{2} + 1.82621 \dots u + 0.41688 \dots

Sea

u_{0} = - 5

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = - 3.39285 \dots : Δ u_{1} = 1.60714 \dots

f (u_{0}) = 2 (- 5)^{3} + 0.91310 \dots (- 5)^{2} + 0.41688 \dots (- 5) + 2.19032 \dots = - 227.06644 \dots

f^{^{'}} (u_{0}) = 6 (- 5)^{2} + 1.82621 \dots (- 5) + 0.41688 \dots = 141.28582 \dots

u_{1} = - 3.39285 \dots

Δ u_{1} = ∣ - 3.39285 \dots - (- 5) ∣ = 1.60714 \dots

Δ u_{1} = 1.60714 \dots

u_{2} = - 2.33697 \dots : Δ u_{2} = 1.05588 \dots

f (u_{1}) = 2 (- 3.39285 \dots)^{3} + 0.91310 \dots (- 3.39285 \dots)^{2} + 0.41688 \dots (- 3.39285 \dots) + 2.19032 \dots = - 66.82653 \dots

f^{^{'}} (u_{1}) = 6 (- 3.39285 \dots)^{2} + 1.82621 \dots (- 3.39285 \dots) + 0.41688 \dots = 63.28970 \dots

u_{2} = - 2.33697 \dots

Δ u_{2} = ∣ - 2.33697 \dots - (- 3.39285 \dots) ∣ = 1.05588 \dots

Δ u_{2} = 1.05588 \dots

u_{3} = - 1.66874 \dots : Δ u_{3} = 0.66822 \dots

f (u_{2}) = 2 (- 2.33697 \dots)^{3} + 0.91310 \dots (- 2.33697 \dots)^{2} + 0.41688 \dots (- 2.33697 \dots) + 2.19032 \dots = - 19.32356 \dots

f^{^{'}} (u_{2}) = 6 (- 2.33697 \dots)^{2} + 1.82621 \dots (- 2.33697 \dots) + 0.41688 \dots = 28.91776 \dots

u_{3} = - 1.66874 \dots

Δ u_{3} = ∣ - 1.66874 \dots - (- 2.33697 \dots) ∣ = 0.66822 \dots

Δ u_{3} = 0.66822 \dots

u_{4} = - 1.29535 \dots : Δ u_{4} = 0.37339 \dots

f (u_{3}) = 2 (- 1.66874 \dots)^{3} + 0.91310 \dots (- 1.66874 \dots)^{2} + 0.41688 \dots (- 1.66874 \dots) + 2.19032 \dots = - 5.25661 \dots

f^{^{'}} (u_{3}) = 6 (- 1.66874 \dots)^{2} + 1.82621 \dots (- 1.66874 \dots) + 0.41688 \dots = 14.07774 \dots

u_{4} = - 1.29535 \dots

Δ u_{4} = ∣ - 1.29535 \dots - (- 1.66874 \dots) ∣ = 0.37339 \dots

Δ u_{4} = 0.37339 \dots

u_{5} = - 1.15191 \dots : Δ u_{5} = 0.14343 \dots

f (u_{4}) = 2 (- 1.29535 \dots)^{3} + 0.91310 \dots (- 1.29535 \dots)^{2} + 0.41688 \dots (- 1.29535 \dots) + 2.19032 \dots = - 1.16457 \dots

f^{^{'}} (u_{4}) = 6 (- 1.29535 \dots)^{2} + 1.82621 \dots (- 1.29535 \dots) + 0.41688 \dots = 8.11890 \dots

u_{5} = - 1.15191 \dots

Δ u_{5} = ∣ - 1.15191 \dots - (- 1.29535 \dots) ∣ = 0.14343 \dots

Δ u_{5} = 0.14343 \dots

u_{6} = - 1.13036 \dots : Δ u_{6} = 0.02155 \dots

f (u_{5}) = 2 (- 1.15191 \dots)^{3} + 0.91310 \dots (- 1.15191 \dots)^{2} + 0.41688 \dots (- 1.15191 \dots) + 2.19032 \dots = - 0.13522 \dots

f^{^{'}} (u_{5}) = 6 (- 1.15191 \dots)^{2} + 1.82621 \dots (- 1.15191 \dots) + 0.41688 \dots = 6.27464 \dots

u_{6} = - 1.13036 \dots

Δ u_{6} = ∣ - 1.13036 \dots - (- 1.15191 \dots) ∣ = 0.02155 \dots

Δ u_{6} = 0.02155 \dots

u_{7} = - 1.12990 \dots : Δ u_{7} = 0.00045 \dots

f (u_{6}) = 2 (- 1.13036 \dots)^{3} + 0.91310 \dots (- 1.13036 \dots)^{2} + 0.41688 \dots (- 1.13036 \dots) + 2.19032 \dots = - 0.00276 \dots

f^{^{'}} (u_{6}) = 6 (- 1.13036 \dots)^{2} + 1.82621 \dots (- 1.13036 \dots) + 0.41688 \dots = 6.01889 \dots

u_{7} = - 1.12990 \dots

Δ u_{7} = ∣ - 1.12990 \dots - (- 1.13036 \dots) ∣ = 0.00045 \dots

Δ u_{7} = 0.00045 \dots

u_{8} = - 1.12990 \dots : Δ u_{8} = 2.06048 E - 7

f (u_{7}) = 2 (- 1.12990 \dots)^{3} + 0.91310 \dots (- 1.12990 \dots)^{2} + 0.41688 \dots (- 1.12990 \dots) + 2.19032 \dots = - 1.23907 E - 6

f^{^{'}} (u_{7}) = 6 (- 1.12990 \dots)^{2} + 1.82621 \dots (- 1.12990 \dots) + 0.41688 \dots = 6.01350 \dots

u_{8} = - 1.12990 \dots

Δ u_{8} = ∣ - 1.12990 \dots - (- 1.12990 \dots) ∣ = 2.06048 E - 7

Δ u_{8} = 2.06048 E - 7

u \approx - 1.12990 \dots

Aplicar la división larga

Eq u a t i o n 0 : \frac{2 u ^{3} + 0.91310 \dots u ^{2} + 0.41688 \dots u + 2.19032 \dots}{u + 1.12990 \dots} = 2 u^{2} - 1.34669 \dots u + 1.93851 \dots

2 u^{2} - 1.34669 \dots u + 1.93851 \dots \approx 0

Encontrar una solución para

2 u^{2} - 1.34669 \dots u + 1.93851 \dots = 0

utilizando el método de Newton-Raphson:

Sin solución para

u \in R

2 u^{2} - 1.34669 \dots u + 1.93851 \dots = 0

Definición del método de Newton-Raphson

f (u) = 2 u^{2} - 1.34669 \dots u + 1.93851 \dots

Hallar

f^{^{'}} (u) : 4 u - 1.34669 \dots

\frac{d}{d u} (2 u^{2} - 1.34669 \dots u + 1.93851 \dots)

Aplicar la regla de la suma/diferencia:

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

= \frac{d}{d u} (2 u^{2}) - \frac{d}{d u} (1.34669 \dots u) + \frac{d}{d u} (1.93851 \dots)

\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}{d u} (1.34669 \dots u) = 1.34669 \dots

\frac{d}{d u} (1.34669 \dots u)

Sacar la constante:

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

= 1.34669 \dots \frac{d u}{d u}

Aplicar la regla de derivación:

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

= 1.34669 \dots \cdot 1

Simplificar

= 1.34669 \dots

\frac{d}{d u} (1.93851 \dots) = 0

\frac{d}{d u} (1.93851 \dots)

Derivada de una constante:

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

= 0

= 4 u - 1.34669 \dots + 0

Simplificar

= 4 u - 1.34669 \dots

Sea

u_{0} = 1

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = 0.02317 \dots : Δ u_{1} = 0.97682 \dots

f (u_{0}) = 2 \cdot 1^{2} - 1.34669 \dots \cdot 1 + 1.93851 \dots = 2.59181 \dots

f^{^{'}} (u_{0}) = 4 \cdot 1 - 1.34669 \dots = 2.65330 \dots

u_{1} = 0.02317 \dots

Δ u_{1} = ∣ 0.02317 \dots - 1 ∣ = 0.97682 \dots

Δ u_{1} = 0.97682 \dots

u_{2} = 1.54500 \dots : Δ u_{2} = 1.52183 \dots

f (u_{1}) = 2 \cdot 0.02317 \dots^{2} - 1.34669 \dots \cdot 0.02317 \dots + 1.93851 \dots = 1.90837 \dots

f^{^{'}} (u_{1}) = 4 \cdot 0.02317 \dots - 1.34669 \dots = - 1.25400 \dots

u_{2} = 1.54500 \dots

Δ u_{2} = ∣ 1.54500 \dots - 0.02317 \dots ∣ = 1.52183 \dots

Δ u_{2} = 1.52183 \dots

u_{3} = 0.58667 \dots : Δ u_{3} = 0.95833 \dots

f (u_{2}) = 2 \cdot 1.54500 \dots^{2} - 1.34669 \dots \cdot 1.54500 \dots + 1.93851 \dots = 4.63194 \dots

f^{^{'}} (u_{2}) = 4 \cdot 1.54500 \dots - 1.34669 \dots = 4.83332 \dots

u_{3} = 0.58667 \dots

Δ u_{3} = ∣ 0.58667 \dots - 1.54500 \dots ∣ = 0.95833 \dots

Δ u_{3} = 0.95833 \dots

u_{4} = - 1.25016 \dots : Δ u_{4} = 1.83683 \dots

f (u_{3}) = 2 \cdot 0.58667 \dots^{2} - 1.34669 \dots \cdot 0.58667 \dots + 1.93851 \dots = 1.83681 \dots

f^{^{'}} (u_{3}) = 4 \cdot 0.58667 \dots - 1.34669 \dots = 0.99998 \dots

u_{4} = - 1.25016 \dots

Δ u_{4} = ∣ - 1.25016 \dots - 0.58667 \dots ∣ = 1.83683 \dots

Δ u_{4} = 1.83683 \dots

u_{5} = - 0.18705 \dots : Δ u_{5} = 1.06311 \dots

f (u_{4}) = 2 (- 1.25016 \dots)^{2} - 1.34669 \dots (- 1.25016 \dots) + 1.93851 \dots = 6.74795 \dots

f^{^{'}} (u_{4}) = 4 (- 1.25016 \dots) - 1.34669 \dots = - 6.34736 \dots

u_{5} = - 0.18705 \dots

Δ u_{5} = ∣ - 0.18705 \dots - (- 1.25016 \dots) ∣ = 1.06311 \dots

Δ u_{5} = 1.06311 \dots

u_{6} = 0.89193 \dots : Δ u_{6} = 1.07898 \dots

f (u_{5}) = 2 (- 0.18705 \dots)^{2} - 1.34669 \dots (- 0.18705 \dots) + 1.93851 \dots = 2.26040 \dots

f^{^{'}} (u_{5}) = 4 (- 0.18705 \dots) - 1.34669 \dots = - 2.09492 \dots

u_{6} = 0.89193 \dots

Δ u_{6} = ∣ 0.89193 \dots - (- 0.18705 \dots) ∣ = 1.07898 \dots

Δ u_{6} = 1.07898 \dots

u_{7} = - 0.15642 \dots : Δ u_{7} = 1.04835 \dots

f (u_{6}) = 2 \cdot 0.89193 \dots^{2} - 1.34669 \dots \cdot 0.89193 \dots + 1.93851 \dots = 2.32843 \dots

f^{^{'}} (u_{6}) = 4 \cdot 0.89193 \dots - 1.34669 \dots = 2.22102 \dots

u_{7} = - 0.15642 \dots

Δ u_{7} = ∣ - 0.15642 \dots - 0.89193 \dots ∣ = 1.04835 \dots

Δ u_{7} = 1.04835 \dots

u_{8} = 0.95800 \dots : Δ u_{8} = 1.11443 \dots

f (u_{7}) = 2 (- 0.15642 \dots)^{2} - 1.34669 \dots (- 0.15642 \dots) + 1.93851 \dots = 2.19811 \dots

f^{^{'}} (u_{7}) = 4 (- 0.15642 \dots) - 1.34669 \dots = - 1.97241 \dots

u_{8} = 0.95800 \dots

Δ u_{8} = ∣ 0.95800 \dots - (- 0.15642 \dots) ∣ = 1.11443 \dots

Δ u_{8} = 1.11443 \dots

No se puede encontrar solución

Las soluciones son

u \approx 0.45655 \dots, u \approx - 1.12990 \dots

Sustituir en la ecuación

u = cos (x)

cos (x) \approx 0.45655 \dots, cos (x) \approx - 1.12990 \dots

cos (x) \approx 0.45655 \dots, cos (x) \approx - 1.12990 \dots

cos (x) = 0.45655 \dots : x = arccos (0.45655 \dots) + 2 πn, x = 2 π - arccos (0.45655 \dots) + 2 πn

cos (x) = 0.45655 \dots

Aplicar propiedades trigonométricas inversas

cos (x) = 0.45655 \dots

Soluciones generales para

cos (x) = 0.45655 \dots

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (0.45655 \dots) + 2 πn, x = 2 π - arccos (0.45655 \dots) + 2 πn

x = arccos (0.45655 \dots) + 2 πn, x = 2 π - arccos (0.45655 \dots) + 2 πn

cos (x) = - 1.12990 \dots :

Sin solución

cos (x) = - 1.12990 \dots

- 1 \leq cos (x) \leq 1

Sin soluci \overset{o}{ˊ} n

Combinar toda las soluciones

x = arccos (0.45655 \dots) + 2 πn, x = 2 π - arccos (0.45655 \dots) + 2 πn

Mostrar soluciones en forma decimal

x = 1.09667 \dots + 2 πn, x = 2 π - 1.09667 \dots + 2 πn

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