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8sin^2(x)2cos(x)=7,0<= x<= 2pi

Solución

8 sin^{2} (x) 2 cos (x) = 7, 0 \leq x \leq 2 π

Solución

Sin soluci \overset{o}{ˊ} n para x \in R

Pasos de solución

8 sin^{2} (x) \cdot 2 cos (x) = 7, 0 \leq x \leq 2 π

Restar

7

de ambos lados

16 sin^{2} (x) cos (x) - 7 = 0

Re-escribir usando identidades trigonométricas

- 7 + 16 cos (x) sin^{2} (x)

Utilizar la identidad pitagórica:

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

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

= - 7 + 16 cos (x) (1 - cos^{2} (x))

- 7 + (1 - cos^{2} (x)) \cdot 16 cos (x) = 0

Usando el método de sustitución

- 7 + (1 - cos^{2} (x)) \cdot 16 cos (x) = 0

Sea:

cos (x) = u

- 7 + (1 - u^{2}) \cdot 16 u = 0

- 7 + (1 - u^{2}) \cdot 16 u = 0 : u \approx - 1.17189 \dots

- 7 + (1 - u^{2}) \cdot 16 u = 0

Desarrollar

- 7 + (1 - u^{2}) \cdot 16 u : - 7 + 16 u - 16 u^{3}

- 7 + (1 - u^{2}) \cdot 16 u

= - 7 + 16 u (1 - u^{2})

Expandir

16 u (1 - u^{2}) : 16 u - 16 u^{3}

16 u (1 - u^{2})

Poner los parentesis utilizando:

a (b - c) = ab - a c

a = 16 u, b = 1, c = u^{2}

= 16 u \cdot 1 - 16 u u^{2}

= 16 \cdot 1 \cdot u - 16 u^{2} u

Simplificar

16 \cdot 1 \cdot u - 16 u^{2} u : 16 u - 16 u^{3}

16 \cdot 1 \cdot u - 16 u^{2} u

16 \cdot 1 \cdot u = 16 u

16 \cdot 1 \cdot u

Multiplicar los numeros:

16 \cdot 1 = 16

= 16 u

16 u^{2} u = 16 u^{3}

16 u^{2} u

Aplicar las leyes de los exponentes:

a^{b} \cdot a^{c} = a^{b + c}

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

= 16 u^{2 + 1}

Sumar:

2 + 1 = 3

= 16 u^{3}

= 16 u - 16 u^{3}

= 16 u - 16 u^{3}

= - 7 + 16 u - 16 u^{3}

- 7 + 16 u - 16 u^{3} = 0

Escribir en la forma binómica

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

- 16 u^{3} + 16 u - 7 = 0

Encontrar una solución para

- 16 u^{3} + 16 u - 7 = 0

utilizando el método de Newton-Raphson:

u \approx - 1.17189 \dots

- 16 u^{3} + 16 u - 7 = 0

Definición del método de Newton-Raphson

f (u) = - 16 u^{3} + 16 u - 7

Hallar

f^{^{'}} (u) : - 48 u^{2} + 16

\frac{d}{d u} (- 16 u^{3} + 16 u - 7)

Aplicar la regla de la suma/diferencia:

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

= - \frac{d}{d u} (16 u^{3}) + \frac{d}{d u} (16 u) - \frac{d}{d u} (7)

\frac{d}{d u} (16 u^{3}) = 48 u^{2}

\frac{d}{d u} (16 u^{3})

Sacar la constante:

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

= 16 \frac{d}{d u} (u^{3})

Aplicar la regla de la potencia:

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

= 16 \cdot 3 u^{3 - 1}

Simplificar

= 48 u^{2}

\frac{d}{d u} (16 u) = 16

\frac{d}{d u} (16 u)

Sacar la constante:

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

= 16 \frac{d u}{d u}

Aplicar la regla de derivación:

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

= 16 \cdot 1

Simplificar

= 16

\frac{d}{d u} (7) = 0

\frac{d}{d u} (7)

Derivada de una constante:

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

= 0

= - 48 u^{2} + 16 - 0

Simplificar

= - 48 u^{2} + 16

Sea

u_{0} = - 2

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = - 1.49431 \dots : Δ u_{1} = 0.50568 \dots

f (u_{0}) = - 16 (- 2)^{3} + 16 (- 2) - 7 = 89

f^{^{'}} (u_{0}) = - 48 (- 2)^{2} + 16 = - 176

u_{1} = - 1.49431 \dots

Δ u_{1} = ∣ - 1.49431 \dots - (- 2) ∣ = 0.50568 \dots

Δ u_{1} = 0.50568 \dots

u_{2} = - 1.24778 \dots : Δ u_{2} = 0.24653 \dots

f (u_{1}) = - 16 (- 1.49431 \dots)^{3} + 16 (- 1.49431 \dots) - 7 = 22.47959 \dots

f^{^{'}} (u_{1}) = - 48 (- 1.49431 \dots)^{2} + 16 = - 91.18336 \dots

u_{2} = - 1.24778 \dots

Δ u_{2} = ∣ - 1.24778 \dots - (- 1.49431 \dots) ∣ = 0.24653 \dots

Δ u_{2} = 0.24653 \dots

u_{3} = - 1.17764 \dots : Δ u_{3} = 0.07014 \dots

f (u_{2}) = - 16 (- 1.24778 \dots)^{3} + 16 (- 1.24778 \dots) - 7 = 4.11969 \dots

f^{^{'}} (u_{2}) = - 48 (- 1.24778 \dots)^{2} + 16 = - 58.73460 \dots

u_{3} = - 1.17764 \dots

Δ u_{3} = ∣ - 1.17764 \dots - (- 1.24778 \dots) ∣ = 0.07014 \dots

Δ u_{3} = 0.07014 \dots

u_{4} = - 1.17192 \dots : Δ u_{4} = 0.00571 \dots

f (u_{3}) = - 16 (- 1.17764 \dots)^{3} + 16 (- 1.17764 \dots) - 7 = 0.28914 \dots

f^{^{'}} (u_{3}) = - 48 (- 1.17764 \dots)^{2} + 16 = - 50.56876 \dots

u_{4} = - 1.17192 \dots

Δ u_{4} = ∣ - 1.17192 \dots - (- 1.17764 \dots) ∣ = 0.00571 \dots

Δ u_{4} = 0.00571 \dots

u_{5} = - 1.17189 \dots : Δ u_{5} = 0.00003 \dots

f (u_{4}) = - 16 (- 1.17192 \dots)^{3} + 16 (- 1.17192 \dots) - 7 = 0.00184 \dots

f^{^{'}} (u_{4}) = - 48 (- 1.17192 \dots)^{2} + 16 = - 49.92391 \dots

u_{5} = - 1.17189 \dots

Δ u_{5} = ∣ - 1.17189 \dots - (- 1.17192 \dots) ∣ = 0.00003 \dots

Δ u_{5} = 0.00003 \dots

u_{6} = - 1.17189 \dots : Δ u_{6} = 1.53907 E - 9

f (u_{5}) = - 16 (- 1.17189 \dots)^{3} + 16 (- 1.17189 \dots) - 7 = 7.68298 E - 8

f^{^{'}} (u_{5}) = - 48 (- 1.17189 \dots)^{2} + 16 = - 49.91975 \dots

u_{6} = - 1.17189 \dots

Δ u_{6} = ∣ - 1.17189 \dots - (- 1.17189 \dots) ∣ = 1.53907 E - 9

Δ u_{6} = 1.53907 E - 9

u \approx - 1.17189 \dots

Aplicar la división larga

Eq u a t i o n 0 : \frac{- 16 u ^{3} + 16 u - 7}{u + 1.17189 \dots} = - 16 u^{2} + 18.75025 \dots u - 5.97325 \dots

- 16 u^{2} + 18.75025 \dots u - 5.97325 \dots \approx 0

Encontrar una solución para

- 16 u^{2} + 18.75025 \dots u - 5.97325 \dots = 0

utilizando el método de Newton-Raphson:

Sin solución para

u \in R

- 16 u^{2} + 18.75025 \dots u - 5.97325 \dots = 0

Definición del método de Newton-Raphson

f (u) = - 16 u^{2} + 18.75025 \dots u - 5.97325 \dots

Hallar

f^{^{'}} (u) : - 32 u + 18.75025 \dots

\frac{d}{d u} (- 16 u^{2} + 18.75025 \dots u - 5.97325 \dots)

Aplicar la regla de la suma/diferencia:

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

= - \frac{d}{d u} (16 u^{2}) + \frac{d}{d u} (18.75025 \dots u) - \frac{d}{d u} (5.97325 \dots)

\frac{d}{d u} (16 u^{2}) = 32 u

\frac{d}{d u} (16 u^{2})

Sacar la constante:

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

= 16 \frac{d}{d u} (u^{2})

Aplicar la regla de la potencia:

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

= 16 \cdot 2 u^{2 - 1}

Simplificar

= 32 u

\frac{d}{d u} (18.75025 \dots u) = 18.75025 \dots

\frac{d}{d u} (18.75025 \dots u)

Sacar la constante:

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

= 18.75025 \dots \frac{d u}{d u}

Aplicar la regla de derivación:

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

= 18.75025 \dots \cdot 1

Simplificar

= 18.75025 \dots

\frac{d}{d u} (5.97325 \dots) = 0

\frac{d}{d u} (5.97325 \dots)

Derivada de una constante:

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

= 0

= - 32 u + 18.75025 \dots - 0

Simplificar

= - 32 u + 18.75025 \dots

Sea

u_{0} = 0

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = 0.31856 \dots : Δ u_{1} = 0.31856 \dots

f (u_{0}) = - 16 \cdot 0^{2} + 18.75025 \dots \cdot 0 - 5.97325 \dots = - 5.97325 \dots

f^{^{'}} (u_{0}) = - 32 \cdot 0 + 18.75025 \dots = 18.75025 \dots

u_{1} = 0.31856 \dots

Δ u_{1} = ∣ 0.31856 \dots - 0 ∣ = 0.31856 \dots

Δ u_{1} = 0.31856 \dots

u_{2} = 0.50835 \dots : Δ u_{2} = 0.18978 \dots

f (u_{1}) = - 16 \cdot 0.31856 \dots^{2} + 18.75025 \dots \cdot 0.31856 \dots - 5.97325 \dots = - 1.62378 \dots

f^{^{'}} (u_{1}) = - 32 \cdot 0.31856 \dots + 18.75025 \dots = 8.55604 \dots

u_{2} = 0.50835 \dots

Δ u_{2} = ∣ 0.50835 \dots - 0.31856 \dots ∣ = 0.18978 \dots

Δ u_{2} = 0.18978 \dots

u_{3} = 0.74043 \dots : Δ u_{3} = 0.23208 \dots

f (u_{2}) = - 16 \cdot 0.50835 \dots^{2} + 18.75025 \dots \cdot 0.50835 \dots - 5.97325 \dots = - 0.57627 \dots

f^{^{'}} (u_{2}) = - 32 \cdot 0.50835 \dots + 18.75025 \dots = 2.48302 \dots

u_{3} = 0.74043 \dots

Δ u_{3} = ∣ 0.74043 \dots - 0.50835 \dots ∣ = 0.23208 \dots

Δ u_{3} = 0.23208 \dots

u_{4} = 0.56610 \dots : Δ u_{4} = 0.17432 \dots

f (u_{3}) = - 16 \cdot 0.74043 \dots^{2} + 18.75025 \dots \cdot 0.74043 \dots - 5.97325 \dots = - 0.86181 \dots

f^{^{'}} (u_{3}) = - 32 \cdot 0.74043 \dots + 18.75025 \dots = - 4.94370 \dots

u_{4} = 0.56610 \dots

Δ u_{4} = ∣ 0.56610 \dots - 0.74043 \dots ∣ = 0.17432 \dots

Δ u_{4} = 0.17432 \dots

u_{5} = 1.33215 \dots : Δ u_{5} = 0.76604 \dots

f (u_{4}) = - 16 \cdot 0.56610 \dots^{2} + 18.75025 \dots \cdot 0.56610 \dots - 5.97325 \dots = - 0.48623 \dots

f^{^{'}} (u_{4}) = - 32 \cdot 0.56610 \dots + 18.75025 \dots = 0.63473 \dots

u_{5} = 1.33215 \dots

Δ u_{5} = ∣ 1.33215 \dots - 0.56610 \dots ∣ = 0.76604 \dots

Δ u_{5} = 0.76604 \dots

u_{6} = 0.93895 \dots : Δ u_{6} = 0.39320 \dots

f (u_{5}) = - 16 \cdot 1.33215 \dots^{2} + 18.75025 \dots \cdot 1.33215 \dots - 5.97325 \dots = - 9.38914 \dots

f^{^{'}} (u_{5}) = - 32 \cdot 1.33215 \dots + 18.75025 \dots = - 23.87863 \dots

u_{6} = 0.93895 \dots

Δ u_{6} = ∣ 0.93895 \dots - 1.33215 \dots ∣ = 0.39320 \dots

Δ u_{6} = 0.39320 \dots

u_{7} = 0.71996 \dots : Δ u_{7} = 0.21898 \dots

f (u_{6}) = - 16 \cdot 0.93895 \dots^{2} + 18.75025 \dots \cdot 0.93895 \dots - 5.97325 \dots = - 2.47373 \dots

f^{^{'}} (u_{6}) = - 32 \cdot 0.93895 \dots + 18.75025 \dots = - 11.29614 \dots

u_{7} = 0.71996 \dots

Δ u_{7} = ∣ 0.71996 \dots - 0.93895 \dots ∣ = 0.21898 \dots

Δ u_{7} = 0.21898 \dots

u_{8} = 0.54103 \dots : Δ u_{8} = 0.17892 \dots

f (u_{7}) = - 16 \cdot 0.71996 \dots^{2} + 18.75025 \dots \cdot 0.71996 \dots - 5.97325 \dots = - 0.76730 \dots

f^{^{'}} (u_{7}) = - 32 \cdot 0.71996 \dots + 18.75025 \dots = - 4.28849 \dots

u_{8} = 0.54103 \dots

Δ u_{8} = ∣ 0.54103 \dots - 0.71996 \dots ∣ = 0.17892 \dots

Δ u_{8} = 0.17892 \dots

u_{9} = 0.89748 \dots : Δ u_{9} = 0.35644 \dots

f (u_{8}) = - 16 \cdot 0.54103 \dots^{2} + 18.75025 \dots \cdot 0.54103 \dots - 5.97325 \dots = - 0.51220 \dots

f^{^{'}} (u_{8}) = - 32 \cdot 0.54103 \dots + 18.75025 \dots = 1.43698 \dots

u_{9} = 0.89748 \dots

Δ u_{9} = ∣ 0.89748 \dots - 0.54103 \dots ∣ = 0.35644 \dots

Δ u_{9} = 0.35644 \dots

u_{10} = 0.69357 \dots : Δ u_{10} = 0.20391 \dots

f (u_{9}) = - 16 \cdot 0.89748 \dots^{2} + 18.75025 \dots \cdot 0.89748 \dots - 5.97325 \dots = - 2.03283 \dots

f^{^{'}} (u_{9}) = - 32 \cdot 0.89748 \dots + 18.75025 \dots = - 9.96922 \dots

u_{10} = 0.69357 \dots

Δ u_{10} = ∣ 0.69357 \dots - 0.89748 \dots ∣ = 0.20391 \dots

Δ u_{10} = 0.20391 \dots

u_{11} = 0.50040 \dots : Δ u_{11} = 0.19316 \dots

f (u_{10}) = - 16 \cdot 0.69357 \dots^{2} + 18.75025 \dots \cdot 0.69357 \dots - 5.97325 \dots = - 0.66527 \dots

f^{^{'}} (u_{10}) = - 32 \cdot 0.69357 \dots + 18.75025 \dots = - 3.44406 \dots

u_{11} = 0.50040 \dots

Δ u_{11} = ∣ 0.50040 \dots - 0.69357 \dots ∣ = 0.19316 \dots

Δ u_{11} = 0.19316 \dots

u_{12} = 0.71851 \dots : Δ u_{12} = 0.21810 \dots

f (u_{11}) = - 16 \cdot 0.50040 \dots^{2} + 18.75025 \dots \cdot 0.50040 \dots - 5.97325 \dots = - 0.59700 \dots

f^{^{'}} (u_{11}) = - 32 \cdot 0.50040 \dots + 18.75025 \dots = 2.73724 \dots

u_{12} = 0.71851 \dots

Δ u_{12} = ∣ 0.71851 \dots - 0.50040 \dots ∣ = 0.21810 \dots

Δ u_{12} = 0.21810 \dots

No se puede encontrar solución

La solución es

u \approx - 1.17189 \dots

Sustituir en la ecuación

u = cos (x)

cos (x) \approx - 1.17189 \dots

cos (x) \approx - 1.17189 \dots

cos (x) = - 1.17189 \dots, 0 \leq x \leq 2 π :

Sin solución

cos (x) = - 1.17189 \dots, 0 \leq x \leq 2 π

- 1 \leq cos (x) \leq 1

Sin soluci \overset{o}{ˊ} n

Combinar toda las soluciones

Sin soluci \overset{o}{ˊ} n para x \in R

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