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2sin^2(x)=sec(x)

Solución

2 sin^{2} (x) = sec (x)

Solución

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

Pasos de solución

2 sin^{2} (x) = sec (x)

Elevar al cuadrado ambos lados

(2 sin^{2} (x))^{2} = sec^{2} (x)

Restar

sec^{2} (x)

de ambos lados

4 sin^{4} (x) - sec^{2} (x) = 0

Expresar con seno, coseno

- sec^{2} (x) + 4 sin^{4} (x)

Utilizar la identidad trigonométrica básica:

sec (x) = \frac{1}{cos ( x )}

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

Simplificar

- (\frac{1}{cos ( x )})^{2} + 4 sin^{4} (x) : \frac{- 1 + 4 sin ^{4} ( x ) cos ^{2} ( x )}{cos ^{2} ( x )}

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

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

(\frac{1}{cos ( x )})^{2}

Aplicar las leyes de los exponentes:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Aplicar la regla

1^{a} = 1

1^{2} = 1

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

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

Convertir a fracción:

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

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

Ya que los denominadores son iguales, combinar las fracciones:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

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

\frac{- 1 + 4 cos ^{2} ( x ) sin ^{4} ( x )}{cos ^{2} ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

- 1 + 4 cos^{2} (x) sin^{4} (x) = 0

Factorizar

- 1 + 4 cos^{2} (x) sin^{4} (x) : (2 sin^{2} (x) cos (x) + 1) (2 sin^{2} (x) cos (x) - 1)

- 1 + 4 cos^{2} (x) sin^{4} (x)

Reescribir

- 1 + 4 sin^{4} (x) cos^{2} (x)

como

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

- 1 + 4 sin^{4} (x) cos^{2} (x)

Reescribir

4

como

2^{2}

= - 1 + 2^{2} sin^{4} (x) cos^{2} (x)

Aplicar las leyes de los exponentes:

a^{b c} = (a^{b})^{c}

sin^{4} (x) = (sin^{2} (x))^{2}

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

Aplicar las leyes de los exponentes:

a^{m} b^{m} = (ab)^{m}

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

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

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

Aplicar la siguiente regla para binomios al cuadrado:

x^{2} - y^{2} = (x + y) (x - y)

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

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

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

Resolver cada parte por separado

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

2 sin^{2} (x) cos (x) + 1 = 0 :

Sin solución

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

Re-escribir usando identidades trigonométricas

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

Utilizar la identidad pitagórica:

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

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

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

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

Usando el método de sustitución

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

Sea:

cos (x) = u

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

1 + (1 - u^{2}) \cdot 2 u = 0 : u \approx 1.19148 \dots

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

Desarrollar

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

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

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

Expandir

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

2 u (1 - u^{2})

Poner los parentesis utilizando:

a (b - c) = ab - a c

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

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

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

Simplificar

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

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

2 \cdot 1 \cdot u = 2 u

2 \cdot 1 \cdot u

Multiplicar los numeros:

2 \cdot 1 = 2

= 2 u

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

2 u^{2} u

Aplicar las leyes de los exponentes:

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

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

= 2 u^{2 + 1}

Sumar:

2 + 1 = 3

= 2 u^{3}

= 2 u - 2 u^{3}

= 2 u - 2 u^{3}

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

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

Escribir en la forma binómica

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

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

Encontrar una solución para

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

utilizando el método de Newton-Raphson:

u \approx 1.19148 \dots

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

Definición del método de Newton-Raphson

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

Hallar

f^{^{'}} (u) : - 6 u^{2} + 2

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

Aplicar la regla de la suma/diferencia:

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

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

\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} (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

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

Simplificar

= - 6 u^{2} + 2

Sea

u_{0} = 1

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = 1.25 : Δ u_{1} = 0.25

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

f^{^{'}} (u_{0}) = - 6 \cdot 1^{2} + 2 = - 4

u_{1} = 1.25

Δ u_{1} = ∣ 1.25 - 1 ∣ = 0.25

Δ u_{1} = 0.25

u_{2} = 1.19491 \dots : Δ u_{2} = 0.05508 \dots

f (u_{1}) = - 2 \cdot 1.2 5^{3} + 2 \cdot 1.25 + 1 = - 0.40625

f^{^{'}} (u_{1}) = - 6 \cdot 1.2 5^{2} + 2 = - 7.375

u_{2} = 1.19491 \dots

Δ u_{2} = ∣ 1.19491 \dots - 1.25 ∣ = 0.05508 \dots

Δ u_{2} = 0.05508 \dots

u_{3} = 1.19150 \dots : Δ u_{3} = 0.00341 \dots

f (u_{2}) = - 2 \cdot 1.19491 \dots^{3} + 2 \cdot 1.19491 \dots + 1 = - 0.02242 \dots

f^{^{'}} (u_{2}) = - 6 \cdot 1.19491 \dots^{2} + 2 = - 6.56693 \dots

u_{3} = 1.19150 \dots

Δ u_{3} = ∣ 1.19150 \dots - 1.19491 \dots ∣ = 0.00341 \dots

Δ u_{3} = 0.00341 \dots

u_{4} = 1.19148 \dots : Δ u_{4} = 0.00001 \dots

f (u_{3}) = - 2 \cdot 1.19150 \dots^{3} + 2 \cdot 1.19150 \dots + 1 = - 0.00008 \dots

f^{^{'}} (u_{3}) = - 6 \cdot 1.19150 \dots^{2} + 2 = - 6.51804 \dots

u_{4} = 1.19148 \dots

Δ u_{4} = ∣ 1.19148 \dots - 1.19150 \dots ∣ = 0.00001 \dots

Δ u_{4} = 0.00001 \dots

u_{5} = 1.19148 \dots : Δ u_{5} = 1.80049 E - 10

f (u_{4}) = - 2 \cdot 1.19148 \dots^{3} + 2 \cdot 1.19148 \dots + 1 = - 1.17354 E - 9

f^{^{'}} (u_{4}) = - 6 \cdot 1.19148 \dots^{2} + 2 = - 6.51786 \dots

u_{5} = 1.19148 \dots

Δ u_{5} = ∣ 1.19148 \dots - 1.19148 \dots ∣ = 1.80049 E - 10

Δ u_{5} = 1.80049 E - 10

u \approx 1.19148 \dots

Aplicar la división larga

Eq u a t i o n 0 : \frac{- 2 u ^{3} + 2 u + 1}{u - 1.19148 \dots} = - 2 u^{2} - 2.38297 \dots u - 0.83928 \dots

- 2 u^{2} - 2.38297 \dots u - 0.83928 \dots \approx 0

Encontrar una solución para

- 2 u^{2} - 2.38297 \dots u - 0.83928 \dots = 0

utilizando el método de Newton-Raphson:

Sin solución para

u \in R

- 2 u^{2} - 2.38297 \dots u - 0.83928 \dots = 0

Definición del método de Newton-Raphson

f (u) = - 2 u^{2} - 2.38297 \dots u - 0.83928 \dots

Hallar

f^{^{'}} (u) : - 4 u - 2.38297 \dots

\frac{d}{d u} (- 2 u^{2} - 2.38297 \dots u - 0.83928 \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} (2.38297 \dots u) - \frac{d}{d u} (0.83928 \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} (2.38297 \dots u) = 2.38297 \dots

\frac{d}{d u} (2.38297 \dots u)

Sacar la constante:

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

= 2.38297 \dots \frac{d u}{d u}

Aplicar la regla de derivación:

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

= 2.38297 \dots \cdot 1

Simplificar

= 2.38297 \dots

\frac{d}{d u} (0.83928 \dots) = 0

\frac{d}{d u} (0.83928 \dots)

Derivada de una constante:

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

= 0

= - 4 u - 2.38297 \dots - 0

Simplificar

= - 4 u - 2.38297 \dots

Sea

u_{0} = 0

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = - 0.35220 \dots : Δ u_{1} = 0.35220 \dots

f (u_{0}) = - 2 \cdot 0^{2} - 2.38297 \dots \cdot 0 - 0.83928 \dots = - 0.83928 \dots

f^{^{'}} (u_{0}) = - 4 \cdot 0 - 2.38297 \dots = - 2.38297 \dots

u_{1} = - 0.35220 \dots

Δ u_{1} = ∣ - 0.35220 \dots - 0 ∣ = 0.35220 \dots

Δ u_{1} = 0.35220 \dots

u_{2} = - 0.60687 \dots : Δ u_{2} = 0.25466 \dots

f (u_{1}) = - 2 (- 0.35220 \dots)^{2} - 2.38297 \dots (- 0.35220 \dots) - 0.83928 \dots = - 0.24809 \dots

f^{^{'}} (u_{1}) = - 4 (- 0.35220 \dots) - 2.38297 \dots = - 0.97417 \dots

u_{2} = - 0.60687 \dots

Δ u_{2} = ∣ - 0.60687 \dots - (- 0.35220 \dots) ∣ = 0.25466 \dots

Δ u_{2} = 0.25466 \dots

u_{3} = 2.30769 \dots : Δ u_{3} = 2.91456 \dots

f (u_{2}) = - 2 (- 0.60687 \dots)^{2} - 2.38297 \dots (- 0.60687 \dots) - 0.83928 \dots = - 0.12971 \dots

f^{^{'}} (u_{2}) = - 4 (- 0.60687 \dots) - 2.38297 \dots = 0.04450 \dots

u_{3} = 2.30769 \dots

Δ u_{3} = ∣ 2.30769 \dots - (- 0.60687 \dots) ∣ = 2.91456 \dots

Δ u_{3} = 2.91456 \dots

u_{4} = 0.84482 \dots : Δ u_{4} = 1.46286 \dots

f (u_{3}) = - 2 \cdot 2.30769 \dots^{2} - 2.38297 \dots \cdot 2.30769 \dots - 0.83928 \dots = - 16.98939 \dots

f^{^{'}} (u_{3}) = - 4 \cdot 2.30769 \dots - 2.38297 \dots = - 11.61376 \dots

u_{4} = 0.84482 \dots

Δ u_{4} = ∣ 0.84482 \dots - 2.30769 \dots ∣ = 1.46286 \dots

Δ u_{4} = 1.46286 \dots

u_{5} = 0.1020746657 : Δ u_{5} = 0.74275 \dots

f (u_{4}) = - 2 \cdot 0.84482 \dots^{2} - 2.38297 \dots \cdot 0.84482 \dots - 0.83928 \dots = - 4.27996 \dots

f^{^{'}} (u_{4}) = - 4 \cdot 0.84482 \dots - 2.38297 \dots = - 5.76228 \dots

u_{5} = 0.1020746657

Δ u_{5} = ∣ 0.1020746657 - 0.84482 \dots ∣ = 0.74275 \dots

Δ u_{5} = 0.74275 \dots

u_{6} = - 0.29321 \dots : Δ u_{6} = 0.39529 \dots

f (u_{5}) = - 2 \cdot 0.102074665 7^{2} - 2.38297 \dots \cdot 0.1020746657 - 0.83928 \dots = - 1.10336 \dots

f^{^{'}} (u_{5}) = - 4 \cdot 0.1020746657 - 2.38297 \dots = - 2.79127 \dots

u_{6} = - 0.29321 \dots

Δ u_{6} = ∣ - 0.29321 \dots - 0.1020746657 ∣ = 0.39529 \dots

Δ u_{6} = 0.39529 \dots

u_{7} = - 0.55146 \dots : Δ u_{7} = 0.25824 \dots

f (u_{6}) = - 2 (- 0.29321 \dots)^{2} - 2.38297 \dots (- 0.29321 \dots) - 0.83928 \dots = - 0.31251 \dots

f^{^{'}} (u_{6}) = - 4 (- 0.29321 \dots) - 2.38297 \dots = - 1.21010 \dots

u_{7} = - 0.55146 \dots

Δ u_{7} = ∣ - 0.55146 \dots - (- 0.29321 \dots) ∣ = 0.25824 \dots

Δ u_{7} = 0.25824 \dots

u_{8} = - 1.30459 \dots : Δ u_{8} = 0.75312 \dots

f (u_{7}) = - 2 (- 0.55146 \dots)^{2} - 2.38297 \dots (- 0.55146 \dots) - 0.83928 \dots = - 0.13338 \dots

f^{^{'}} (u_{7}) = - 4 (- 0.55146 \dots) - 2.38297 \dots = - 0.17710 \dots

u_{8} = - 1.30459 \dots

Δ u_{8} = ∣ - 1.30459 \dots - (- 0.55146 \dots) ∣ = 0.75312 \dots

Δ u_{8} = 0.75312 \dots

u_{9} = - 0.90450 \dots : Δ u_{9} = 0.40008 \dots

f (u_{8}) = - 2 (- 1.30459 \dots)^{2} - 2.38297 \dots (- 1.30459 \dots) - 0.83928 \dots = - 1.13440 \dots

f^{^{'}} (u_{8}) = - 4 (- 1.30459 \dots) - 2.38297 \dots = 2.83540 \dots

u_{9} = - 0.90450 \dots

Δ u_{9} = ∣ - 0.90450 \dots - (- 1.30459 \dots) ∣ = 0.40008 \dots

Δ u_{9} = 0.40008 \dots

u_{10} = - 0.64530 \dots : Δ u_{10} = 0.25920 \dots

f (u_{9}) = - 2 (- 0.90450 \dots)^{2} - 2.38297 \dots (- 0.90450 \dots) - 0.83928 \dots = - 0.32013 \dots

f^{^{'}} (u_{9}) = - 4 (- 0.90450 \dots) - 2.38297 \dots = 1.23506 \dots

u_{10} = - 0.64530 \dots

Δ u_{10} = ∣ - 0.64530 \dots - (- 0.90450 \dots) ∣ = 0.25920 \dots

Δ u_{10} = 0.25920 \dots

u_{11} = 0.03257 \dots : Δ u_{11} = 0.67788 \dots

f (u_{10}) = - 2 (- 0.64530 \dots)^{2} - 2.38297 \dots (- 0.64530 \dots) - 0.83928 \dots = - 0.13437 \dots

f^{^{'}} (u_{10}) = - 4 (- 0.64530 \dots) - 2.38297 \dots = 0.19823 \dots

u_{11} = 0.03257 \dots

Δ u_{11} = ∣ 0.03257 \dots - (- 0.64530 \dots) ∣ = 0.67788 \dots

Δ u_{11} = 0.67788 \dots

u_{12} = - 0.33309 \dots : Δ u_{12} = 0.36567 \dots

f (u_{11}) = - 2 \cdot 0.03257 \dots^{2} - 2.38297 \dots \cdot 0.03257 \dots - 0.83928 \dots = - 0.91904 \dots

f^{^{'}} (u_{11}) = - 4 \cdot 0.03257 \dots - 2.38297 \dots = - 2.51329 \dots

u_{12} = - 0.33309 \dots

Δ u_{12} = ∣ - 0.33309 \dots - 0.03257 \dots ∣ = 0.36567 \dots

Δ u_{12} = 0.36567 \dots

u_{13} = - 0.58764 \dots : Δ u_{13} = 0.25455 \dots

f (u_{12}) = - 2 (- 0.33309 \dots)^{2} - 2.38297 \dots (- 0.33309 \dots) - 0.83928 \dots = - 0.26743 \dots

f^{^{'}} (u_{12}) = - 4 (- 0.33309 \dots) - 2.38297 \dots = - 1.05059 \dots

u_{13} = - 0.58764 \dots

Δ u_{13} = ∣ - 0.58764 \dots - (- 0.33309 \dots) ∣ = 0.25455 \dots

Δ u_{13} = 0.25455 \dots

u_{14} = - 4.59022 \dots : Δ u_{14} = 4.00257 \dots

f (u_{13}) = - 2 (- 0.58764 \dots)^{2} - 2.38297 \dots (- 0.58764 \dots) - 0.83928 \dots = - 0.12959 \dots

f^{^{'}} (u_{13}) = - 4 (- 0.58764 \dots) - 2.38297 \dots = - 0.03237 \dots

u_{14} = - 4.59022 \dots

Δ u_{14} = ∣ - 4.59022 \dots - (- 0.58764 \dots) ∣ = 4.00257 \dots

Δ u_{14} = 4.00257 \dots

No se puede encontrar solución

La solución es

u \approx 1.19148 \dots

Sustituir en la ecuación

u = cos (x)

cos (x) \approx 1.19148 \dots

cos (x) \approx 1.19148 \dots

cos (x) = 1.19148 \dots :

Sin solución

cos (x) = 1.19148 \dots

- 1 \leq cos (x) \leq 1

Sin soluci \overset{o}{ˊ} n

Combinar toda las soluciones

Sin soluci \overset{o}{ˊ} n

2 sin^{2} (x) cos (x) - 1 = 0 :

Sin solución

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

Re-escribir usando identidades trigonométricas

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

Utilizar la identidad pitagórica:

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

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

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

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

Usando el método de sustitución

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

Sea:

cos (x) = u

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

- 1 + (1 - u^{2}) \cdot 2 u = 0 : u \approx - 1.19148 \dots

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

Desarrollar

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

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

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

Expandir

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

2 u (1 - u^{2})

Poner los parentesis utilizando:

a (b - c) = ab - a c

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

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

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

Simplificar

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

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

2 \cdot 1 \cdot u = 2 u

2 \cdot 1 \cdot u

Multiplicar los numeros:

2 \cdot 1 = 2

= 2 u

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

2 u^{2} u

Aplicar las leyes de los exponentes:

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

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

= 2 u^{2 + 1}

Sumar:

2 + 1 = 3

= 2 u^{3}

= 2 u - 2 u^{3}

= 2 u - 2 u^{3}

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

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

Escribir en la forma binómica

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

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

Encontrar una solución para

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

utilizando el método de Newton-Raphson:

u \approx - 1.19148 \dots

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

Definición del método de Newton-Raphson

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

Hallar

f^{^{'}} (u) : - 6 u^{2} + 2

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

Aplicar la regla de la suma/diferencia:

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

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

\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} (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

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

Simplificar

= - 6 u^{2} + 2

Sea

u_{0} = - 1

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = - 1.25 : Δ u_{1} = 0.25

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

f^{^{'}} (u_{0}) = - 6 (- 1)^{2} + 2 = - 4

u_{1} = - 1.25

Δ u_{1} = ∣ - 1.25 - (- 1) ∣ = 0.25

Δ u_{1} = 0.25

u_{2} = - 1.19491 \dots : Δ u_{2} = 0.05508 \dots

f (u_{1}) = - 2 (- 1.25)^{3} + 2 (- 1.25) - 1 = 0.40625

f^{^{'}} (u_{1}) = - 6 (- 1.25)^{2} + 2 = - 7.375

u_{2} = - 1.19491 \dots

Δ u_{2} = ∣ - 1.19491 \dots - (- 1.25) ∣ = 0.05508 \dots

Δ u_{2} = 0.05508 \dots

u_{3} = - 1.19150 \dots : Δ u_{3} = 0.00341 \dots

f (u_{2}) = - 2 (- 1.19491 \dots)^{3} + 2 (- 1.19491 \dots) - 1 = 0.02242 \dots

f^{^{'}} (u_{2}) = - 6 (- 1.19491 \dots)^{2} + 2 = - 6.56693 \dots

u_{3} = - 1.19150 \dots

Δ u_{3} = ∣ - 1.19150 \dots - (- 1.19491 \dots) ∣ = 0.00341 \dots

Δ u_{3} = 0.00341 \dots

u_{4} = - 1.19148 \dots : Δ u_{4} = 0.00001 \dots

f (u_{3}) = - 2 (- 1.19150 \dots)^{3} + 2 (- 1.19150 \dots) - 1 = 0.00008 \dots

f^{^{'}} (u_{3}) = - 6 (- 1.19150 \dots)^{2} + 2 = - 6.51804 \dots

u_{4} = - 1.19148 \dots

Δ u_{4} = ∣ - 1.19148 \dots - (- 1.19150 \dots) ∣ = 0.00001 \dots

Δ u_{4} = 0.00001 \dots

u_{5} = - 1.19148 \dots : Δ u_{5} = 1.80049 E - 10

f (u_{4}) = - 2 (- 1.19148 \dots)^{3} + 2 (- 1.19148 \dots) - 1 = 1.17354 E - 9

f^{^{'}} (u_{4}) = - 6 (- 1.19148 \dots)^{2} + 2 = - 6.51786 \dots

u_{5} = - 1.19148 \dots

Δ u_{5} = ∣ - 1.19148 \dots - (- 1.19148 \dots) ∣ = 1.80049 E - 10

Δ u_{5} = 1.80049 E - 10

u \approx - 1.19148 \dots

Aplicar la división larga

Eq u a t i o n 0 : \frac{- 2 u ^{3} + 2 u - 1}{u + 1.19148 \dots} = - 2 u^{2} + 2.38297 \dots u - 0.83928 \dots

- 2 u^{2} + 2.38297 \dots u - 0.83928 \dots \approx 0

Encontrar una solución para

- 2 u^{2} + 2.38297 \dots u - 0.83928 \dots = 0

utilizando el método de Newton-Raphson:

Sin solución para

u \in R

- 2 u^{2} + 2.38297 \dots u - 0.83928 \dots = 0

Definición del método de Newton-Raphson

f (u) = - 2 u^{2} + 2.38297 \dots u - 0.83928 \dots

Hallar

f^{^{'}} (u) : - 4 u + 2.38297 \dots

\frac{d}{d u} (- 2 u^{2} + 2.38297 \dots u - 0.83928 \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} (2.38297 \dots u) - \frac{d}{d u} (0.83928 \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} (2.38297 \dots u) = 2.38297 \dots

\frac{d}{d u} (2.38297 \dots u)

Sacar la constante:

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

= 2.38297 \dots \frac{d u}{d u}

Aplicar la regla de derivación:

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

= 2.38297 \dots \cdot 1

Simplificar

= 2.38297 \dots

\frac{d}{d u} (0.83928 \dots) = 0

\frac{d}{d u} (0.83928 \dots)

Derivada de una constante:

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

= 0

= - 4 u + 2.38297 \dots - 0

Simplificar

= - 4 u + 2.38297 \dots

Sea

u_{0} = 0

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = 0.35220 \dots : Δ u_{1} = 0.35220 \dots

f (u_{0}) = - 2 \cdot 0^{2} + 2.38297 \dots \cdot 0 - 0.83928 \dots = - 0.83928 \dots

f^{^{'}} (u_{0}) = - 4 \cdot 0 + 2.38297 \dots = 2.38297 \dots

u_{1} = 0.35220 \dots

Δ u_{1} = ∣ 0.35220 \dots - 0 ∣ = 0.35220 \dots

Δ u_{1} = 0.35220 \dots

u_{2} = 0.60687 \dots : Δ u_{2} = 0.25466 \dots

f (u_{1}) = - 2 \cdot 0.35220 \dots^{2} + 2.38297 \dots \cdot 0.35220 \dots - 0.83928 \dots = - 0.24809 \dots

f^{^{'}} (u_{1}) = - 4 \cdot 0.35220 \dots + 2.38297 \dots = 0.97417 \dots

u_{2} = 0.60687 \dots

Δ u_{2} = ∣ 0.60687 \dots - 0.35220 \dots ∣ = 0.25466 \dots

Δ u_{2} = 0.25466 \dots

u_{3} = - 2.30769 \dots : Δ u_{3} = 2.91456 \dots

f (u_{2}) = - 2 \cdot 0.60687 \dots^{2} + 2.38297 \dots \cdot 0.60687 \dots - 0.83928 \dots = - 0.12971 \dots

f^{^{'}} (u_{2}) = - 4 \cdot 0.60687 \dots + 2.38297 \dots = - 0.04450 \dots

u_{3} = - 2.30769 \dots

Δ u_{3} = ∣ - 2.30769 \dots - 0.60687 \dots ∣ = 2.91456 \dots

Δ u_{3} = 2.91456 \dots

u_{4} = - 0.84482 \dots : Δ u_{4} = 1.46286 \dots

f (u_{3}) = - 2 (- 2.30769 \dots)^{2} + 2.38297 \dots (- 2.30769 \dots) - 0.83928 \dots = - 16.98939 \dots

f^{^{'}} (u_{3}) = - 4 (- 2.30769 \dots) + 2.38297 \dots = 11.61376 \dots

u_{4} = - 0.84482 \dots

Δ u_{4} = ∣ - 0.84482 \dots - (- 2.30769 \dots) ∣ = 1.46286 \dots

Δ u_{4} = 1.46286 \dots

u_{5} = - 0.1020746657 : Δ u_{5} = 0.74275 \dots

f (u_{4}) = - 2 (- 0.84482 \dots)^{2} + 2.38297 \dots (- 0.84482 \dots) - 0.83928 \dots = - 4.27996 \dots

f^{^{'}} (u_{4}) = - 4 (- 0.84482 \dots) + 2.38297 \dots = 5.76228 \dots

u_{5} = - 0.1020746657

Δ u_{5} = ∣ - 0.1020746657 - (- 0.84482 \dots) ∣ = 0.74275 \dots

Δ u_{5} = 0.74275 \dots

u_{6} = 0.29321 \dots : Δ u_{6} = 0.39529 \dots

f (u_{5}) = - 2 (- 0.1020746657)^{2} + 2.38297 \dots (- 0.1020746657) - 0.83928 \dots = - 1.10336 \dots

f^{^{'}} (u_{5}) = - 4 (- 0.1020746657) + 2.38297 \dots = 2.79127 \dots

u_{6} = 0.29321 \dots

Δ u_{6} = ∣ 0.29321 \dots - (- 0.1020746657) ∣ = 0.39529 \dots

Δ u_{6} = 0.39529 \dots

u_{7} = 0.55146 \dots : Δ u_{7} = 0.25824 \dots

f (u_{6}) = - 2 \cdot 0.29321 \dots^{2} + 2.38297 \dots \cdot 0.29321 \dots - 0.83928 \dots = - 0.31251 \dots

f^{^{'}} (u_{6}) = - 4 \cdot 0.29321 \dots + 2.38297 \dots = 1.21010 \dots

u_{7} = 0.55146 \dots

Δ u_{7} = ∣ 0.55146 \dots - 0.29321 \dots ∣ = 0.25824 \dots

Δ u_{7} = 0.25824 \dots

u_{8} = 1.30459 \dots : Δ u_{8} = 0.75312 \dots

f (u_{7}) = - 2 \cdot 0.55146 \dots^{2} + 2.38297 \dots \cdot 0.55146 \dots - 0.83928 \dots = - 0.13338 \dots

f^{^{'}} (u_{7}) = - 4 \cdot 0.55146 \dots + 2.38297 \dots = 0.17710 \dots

u_{8} = 1.30459 \dots

Δ u_{8} = ∣ 1.30459 \dots - 0.55146 \dots ∣ = 0.75312 \dots

Δ u_{8} = 0.75312 \dots

u_{9} = 0.90450 \dots : Δ u_{9} = 0.40008 \dots

f (u_{8}) = - 2 \cdot 1.30459 \dots^{2} + 2.38297 \dots \cdot 1.30459 \dots - 0.83928 \dots = - 1.13440 \dots

f^{^{'}} (u_{8}) = - 4 \cdot 1.30459 \dots + 2.38297 \dots = - 2.83540 \dots

u_{9} = 0.90450 \dots

Δ u_{9} = ∣ 0.90450 \dots - 1.30459 \dots ∣ = 0.40008 \dots

Δ u_{9} = 0.40008 \dots

u_{10} = 0.64530 \dots : Δ u_{10} = 0.25920 \dots

f (u_{9}) = - 2 \cdot 0.90450 \dots^{2} + 2.38297 \dots \cdot 0.90450 \dots - 0.83928 \dots = - 0.32013 \dots

f^{^{'}} (u_{9}) = - 4 \cdot 0.90450 \dots + 2.38297 \dots = - 1.23506 \dots

u_{10} = 0.64530 \dots

Δ u_{10} = ∣ 0.64530 \dots - 0.90450 \dots ∣ = 0.25920 \dots

Δ u_{10} = 0.25920 \dots

u_{11} = - 0.03257 \dots : Δ u_{11} = 0.67788 \dots

f (u_{10}) = - 2 \cdot 0.64530 \dots^{2} + 2.38297 \dots \cdot 0.64530 \dots - 0.83928 \dots = - 0.13437 \dots

f^{^{'}} (u_{10}) = - 4 \cdot 0.64530 \dots + 2.38297 \dots = - 0.19823 \dots

u_{11} = - 0.03257 \dots

Δ u_{11} = ∣ - 0.03257 \dots - 0.64530 \dots ∣ = 0.67788 \dots

Δ u_{11} = 0.67788 \dots

u_{12} = 0.33309 \dots : Δ u_{12} = 0.36567 \dots

f (u_{11}) = - 2 (- 0.03257 \dots)^{2} + 2.38297 \dots (- 0.03257 \dots) - 0.83928 \dots = - 0.91904 \dots

f^{^{'}} (u_{11}) = - 4 (- 0.03257 \dots) + 2.38297 \dots = 2.51329 \dots

u_{12} = 0.33309 \dots

Δ u_{12} = ∣ 0.33309 \dots - (- 0.03257 \dots) ∣ = 0.36567 \dots

Δ u_{12} = 0.36567 \dots

u_{13} = 0.58764 \dots : Δ u_{13} = 0.25455 \dots

f (u_{12}) = - 2 \cdot 0.33309 \dots^{2} + 2.38297 \dots \cdot 0.33309 \dots - 0.83928 \dots = - 0.26743 \dots

f^{^{'}} (u_{12}) = - 4 \cdot 0.33309 \dots + 2.38297 \dots = 1.05059 \dots

u_{13} = 0.58764 \dots

Δ u_{13} = ∣ 0.58764 \dots - 0.33309 \dots ∣ = 0.25455 \dots

Δ u_{13} = 0.25455 \dots

u_{14} = 4.59022 \dots : Δ u_{14} = 4.00257 \dots

f (u_{13}) = - 2 \cdot 0.58764 \dots^{2} + 2.38297 \dots \cdot 0.58764 \dots - 0.83928 \dots = - 0.12959 \dots

f^{^{'}} (u_{13}) = - 4 \cdot 0.58764 \dots + 2.38297 \dots = 0.03237 \dots

u_{14} = 4.59022 \dots

Δ u_{14} = ∣ 4.59022 \dots - 0.58764 \dots ∣ = 4.00257 \dots

Δ u_{14} = 4.00257 \dots

No se puede encontrar solución

La solución es

u \approx - 1.19148 \dots

Sustituir en la ecuación

u = cos (x)

cos (x) \approx - 1.19148 \dots

cos (x) \approx - 1.19148 \dots

cos (x) = - 1.19148 \dots :

Sin solución

cos (x) = - 1.19148 \dots

- 1 \leq cos (x) \leq 1

Sin soluci \overset{o}{ˊ} n

Combinar toda las soluciones

Sin soluci \overset{o}{ˊ} n

Combinar toda las soluciones

Sin soluci \overset{o}{ˊ} n

Verificar las soluciones sustituyendo en la ecuación original

Verificar las soluciones sustituyéndolas en

2 sin^{2} (x) = sec (x)

Quitar las que no concuerden con la ecuación.

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

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\frac{3}{cos ^{2} ( x )} = 3 + 2 tan^{2} (x)

13*sin(x)=12

13 \cdot sin (x) = 12

11sin(θ)=11cos(θ)

11 sin (θ) = 11 cos (θ)