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2cos^2(t)-8sin^2(t)cos(t)-1=0

Solución

2 cos^{2} (t) - 8 sin^{2} (t) cos (t) - 1 = 0

Solución

t = 1.69418 \dots + 2 πn, t = - 1.69418 \dots + 2 πn, t = 0.32915 \dots + 2 πn, t = 2 π - 0.32915 \dots + 2 πn

Grados

t = 97.06973 \dots^{\circ} + 36 0^{\circ} n, t = - 97.06973 \dots^{\circ} + 36 0^{\circ} n, t = 18.85929 \dots^{\circ} + 36 0^{\circ} n, t = 341.14070 \dots^{\circ} + 36 0^{\circ} n

Pasos de solución

2 cos^{2} (t) - 8 sin^{2} (t) cos (t) - 1 = 0

Re-escribir usando identidades trigonométricas

- 1 + 2 cos^{2} (t) - 8 cos (t) sin^{2} (t)

Utilizar la identidad pitagórica:

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

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

= - 1 + 2 cos^{2} (t) - 8 cos (t) (1 - cos^{2} (t))

- 1 + 2 cos^{2} (t) - (1 - cos^{2} (t)) \cdot 8 cos (t) = 0

Usando el método de sustitución

- 1 + 2 cos^{2} (t) - (1 - cos^{2} (t)) \cdot 8 cos (t) = 0

Sea:

cos (t) = u

- 1 + 2 u^{2} - (1 - u^{2}) \cdot 8 u = 0

- 1 + 2 u^{2} - (1 - u^{2}) \cdot 8 u = 0 : u \approx - 0.12307 \dots, u \approx 0.94631 \dots, u \approx - 1.07323 \dots

- 1 + 2 u^{2} - (1 - u^{2}) \cdot 8 u = 0

Desarrollar

- 1 + 2 u^{2} - (1 - u^{2}) \cdot 8 u : - 1 + 2 u^{2} - 8 u + 8 u^{3}

- 1 + 2 u^{2} - (1 - u^{2}) \cdot 8 u

= - 1 + 2 u^{2} - 8 u (1 - u^{2})

Expandir

- 8 u (1 - u^{2}) : - 8 u + 8 u^{3}

- 8 u (1 - u^{2})

Poner los parentesis utilizando:

a (b - c) = ab - a c

a = - 8 u, b = 1, c = u^{2}

= - 8 u \cdot 1 - (- 8 u) u^{2}

Aplicar las reglas de los signos

- (- a) = a

= - 8 \cdot 1 \cdot u + 8 u^{2} u

Simplificar

- 8 \cdot 1 \cdot u + 8 u^{2} u : - 8 u + 8 u^{3}

- 8 \cdot 1 \cdot u + 8 u^{2} u

8 \cdot 1 \cdot u = 8 u

8 \cdot 1 \cdot u

Multiplicar los numeros:

8 \cdot 1 = 8

= 8 u

8 u^{2} u = 8 u^{3}

8 u^{2} u

Aplicar las leyes de los exponentes:

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

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

= 8 u^{2 + 1}

Sumar:

2 + 1 = 3

= 8 u^{3}

= - 8 u + 8 u^{3}

= - 8 u + 8 u^{3}

= - 1 + 2 u^{2} - 8 u + 8 u^{3}

- 1 + 2 u^{2} - 8 u + 8 u^{3} = 0

Escribir en la forma binómica

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

8 u^{3} + 2 u^{2} - 8 u - 1 = 0

Encontrar una solución para

8 u^{3} + 2 u^{2} - 8 u - 1 = 0

utilizando el método de Newton-Raphson:

u \approx - 0.12307 \dots

8 u^{3} + 2 u^{2} - 8 u - 1 = 0

Definición del método de Newton-Raphson

f (u) = 8 u^{3} + 2 u^{2} - 8 u - 1

Hallar

f^{^{'}} (u) : 24 u^{2} + 4 u - 8

\frac{d}{d u} (8 u^{3} + 2 u^{2} - 8 u - 1)

Aplicar la regla de la suma/diferencia:

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

= \frac{d}{d u} (8 u^{3}) + \frac{d}{d u} (2 u^{2}) - \frac{d}{d u} (8 u) - \frac{d}{d u} (1)

\frac{d}{d u} (8 u^{3}) = 24 u^{2}

\frac{d}{d u} (8 u^{3})

Sacar la constante:

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

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

Aplicar la regla de la potencia:

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

= 8 \cdot 3 u^{3 - 1}

Simplificar

= 24 u^{2}

\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} (8 u) = 8

\frac{d}{d u} (8 u)

Sacar la constante:

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

= 8 \frac{d u}{d u}

Aplicar la regla de derivación:

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

= 8 \cdot 1

Simplificar

= 8

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

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

Derivada de una constante:

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

= 0

= 24 u^{2} + 4 u - 8 - 0

Simplificar

= 24 u^{2} + 4 u - 8

Sea

u_{0} = 0

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = - 0.125 : Δ u_{1} = 0.125

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

f^{^{'}} (u_{0}) = 24 \cdot 0^{2} + 4 \cdot 0 - 8 = - 8

u_{1} = - 0.125

Δ u_{1} = ∣ - 0.125 - 0 ∣ = 0.125

Δ u_{1} = 0.125

u_{2} = - 0.12307 \dots : Δ u_{2} = 0.00192 \dots

f (u_{1}) = 8 (- 0.125)^{3} + 2 (- 0.125)^{2} - 8 (- 0.125) - 1 = 0.015625

f^{^{'}} (u_{1}) = 24 (- 0.125)^{2} + 4 (- 0.125) - 8 = - 8.125

u_{2} = - 0.12307 \dots

Δ u_{2} = ∣ - 0.12307 \dots - (- 0.125) ∣ = 0.00192 \dots

Δ u_{2} = 0.00192 \dots

u_{3} = - 0.12307 \dots : Δ u_{3} = 4.47956 E - 7

f (u_{2}) = 8 (- 0.12307 \dots)^{3} + 2 (- 0.12307 \dots)^{2} - 8 (- 0.12307 \dots) - 1 = - 3.64133 E - 6

f^{^{'}} (u_{2}) = 24 (- 0.12307 \dots)^{2} + 4 (- 0.12307 \dots) - 8 = - 8.12875 \dots

u_{3} = - 0.12307 \dots

Δ u_{3} = ∣ - 0.12307 \dots - (- 0.12307 \dots) ∣ = 4.47956 E - 7

Δ u_{3} = 4.47956 E - 7

u \approx - 0.12307 \dots

Aplicar la división larga

Eq u a t i o n 0 : \frac{8 u ^{3} + 2 u ^{2} - 8 u - 1}{u + 0.12307 \dots} = 8 u^{2} + 1.01538 \dots u - 8.12497 \dots

8 u^{2} + 1.01538 \dots u - 8.12497 \dots \approx 0

Encontrar una solución para

8 u^{2} + 1.01538 \dots u - 8.12497 \dots = 0

utilizando el método de Newton-Raphson:

u \approx 0.94631 \dots

8 u^{2} + 1.01538 \dots u - 8.12497 \dots = 0

Definición del método de Newton-Raphson

f (u) = 8 u^{2} + 1.01538 \dots u - 8.12497 \dots

Hallar

f^{^{'}} (u) : 16 u + 1.01538 \dots

\frac{d}{d u} (8 u^{2} + 1.01538 \dots u - 8.12497 \dots)

Aplicar la regla de la suma/diferencia:

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

= \frac{d}{d u} (8 u^{2}) + \frac{d}{d u} (1.01538 \dots u) - \frac{d}{d u} (8.12497 \dots)

\frac{d}{d u} (8 u^{2}) = 16 u

\frac{d}{d u} (8 u^{2})

Sacar la constante:

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

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

Aplicar la regla de la potencia:

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

= 8 \cdot 2 u^{2 - 1}

Simplificar

= 16 u

\frac{d}{d u} (1.01538 \dots u) = 1.01538 \dots

\frac{d}{d u} (1.01538 \dots u)

Sacar la constante:

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

= 1.01538 \dots \frac{d u}{d u}

Aplicar la regla de derivación:

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

= 1.01538 \dots \cdot 1

Simplificar

= 1.01538 \dots

\frac{d}{d u} (8.12497 \dots) = 0

\frac{d}{d u} (8.12497 \dots)

Derivada de una constante:

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

= 0

= 16 u + 1.01538 \dots - 0

Simplificar

= 16 u + 1.01538 \dots

Sea

u_{0} = 5

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = 2.56895 \dots : Δ u_{1} = 2.43104 \dots

f (u_{0}) = 8 \cdot 5^{2} + 1.01538 \dots \cdot 5 - 8.12497 \dots = 196.95193 \dots

f^{^{'}} (u_{0}) = 16 \cdot 5 + 1.01538 \dots = 81.01538 \dots

u_{1} = 2.56895 \dots

Δ u_{1} = ∣ 2.56895 \dots - 5 ∣ = 2.43104 \dots

Δ u_{1} = 2.43104 \dots

u_{2} = 1.44641 \dots : Δ u_{2} = 1.12253 \dots

f (u_{1}) = 8 \cdot 2.56895 \dots^{2} + 1.01538 \dots \cdot 2.56895 \dots - 8.12497 \dots = 47.27978 \dots

f^{^{'}} (u_{1}) = 16 \cdot 2.56895 \dots + 1.01538 \dots = 42.11868 \dots

u_{2} = 1.44641 \dots

Δ u_{2} = ∣ 1.44641 \dots - 2.56895 \dots ∣ = 1.12253 \dots

Δ u_{2} = 1.12253 \dots

u_{3} = 1.02913 \dots : Δ u_{3} = 0.41728 \dots

f (u_{2}) = 8 \cdot 1.44641 \dots^{2} + 1.01538 \dots \cdot 1.44641 \dots - 8.12497 \dots = 10.08071 \dots

f^{^{'}} (u_{2}) = 16 \cdot 1.44641 \dots + 1.01538 \dots = 24.15808 \dots

u_{3} = 1.02913 \dots

Δ u_{3} = ∣ 1.02913 \dots - 1.44641 \dots ∣ = 0.41728 \dots

Δ u_{3} = 0.41728 \dots

u_{4} = 0.94945 \dots : Δ u_{4} = 0.07968 \dots

f (u_{3}) = 8 \cdot 1.02913 \dots^{2} + 1.01538 \dots \cdot 1.02913 \dots - 8.12497 \dots = 1.39299 \dots

f^{^{'}} (u_{3}) = 16 \cdot 1.02913 \dots + 1.01538 \dots = 17.48158 \dots

u_{4} = 0.94945 \dots

Δ u_{4} = ∣ 0.94945 \dots - 1.02913 \dots ∣ = 0.07968 \dots

Δ u_{4} = 0.07968 \dots

u_{5} = 0.94632 \dots : Δ u_{5} = 0.00313 \dots

f (u_{4}) = 8 \cdot 0.94945 \dots^{2} + 1.01538 \dots \cdot 0.94945 \dots - 8.12497 \dots = 0.05079 \dots

f^{^{'}} (u_{4}) = 16 \cdot 0.94945 \dots + 1.01538 \dots = 16.20665 \dots

u_{5} = 0.94632 \dots

Δ u_{5} = ∣ 0.94632 \dots - 0.94945 \dots ∣ = 0.00313 \dots

Δ u_{5} = 0.00313 \dots

u_{6} = 0.94631 \dots : Δ u_{6} = 4.86413 E - 6

f (u_{5}) = 8 \cdot 0.94632 \dots^{2} + 1.01538 \dots \cdot 0.94632 \dots - 8.12497 \dots = 0.00007 \dots

f^{^{'}} (u_{5}) = 16 \cdot 0.94632 \dots + 1.01538 \dots = 16.15650 \dots

u_{6} = 0.94631 \dots

Δ u_{6} = ∣ 0.94631 \dots - 0.94632 \dots ∣ = 4.86413 E - 6

Δ u_{6} = 4.86413 E - 6

u_{7} = 0.94631 \dots : Δ u_{7} = 1.17153 E - 11

f (u_{6}) = 8 \cdot 0.94631 \dots^{2} + 1.01538 \dots \cdot 0.94631 \dots - 8.12497 \dots = 1.89278 E - 10

f^{^{'}} (u_{6}) = 16 \cdot 0.94631 \dots + 1.01538 \dots = 16.15642 \dots

u_{7} = 0.94631 \dots

Δ u_{7} = ∣ 0.94631 \dots - 0.94631 \dots ∣ = 1.17153 E - 11

Δ u_{7} = 1.17153 E - 11

u \approx 0.94631 \dots

Aplicar la división larga

Eq u a t i o n 0 : \frac{8 u ^{2} + 1.01538 \dots u - 8.12497 \dots}{u - 0.94631 \dots} = 8 u + 8.58590 \dots

8 u + 8.58590 \dots \approx 0

u \approx - 1.07323 \dots

Las soluciones son

u \approx - 0.12307 \dots, u \approx 0.94631 \dots, u \approx - 1.07323 \dots

Sustituir en la ecuación

u = cos (t)

cos (t) \approx - 0.12307 \dots, cos (t) \approx 0.94631 \dots, cos (t) \approx - 1.07323 \dots

cos (t) \approx - 0.12307 \dots, cos (t) \approx 0.94631 \dots, cos (t) \approx - 1.07323 \dots

cos (t) = - 0.12307 \dots : t = arccos (- 0.12307 \dots) + 2 πn, t = - arccos (- 0.12307 \dots) + 2 πn

cos (t) = - 0.12307 \dots

Aplicar propiedades trigonométricas inversas

cos (t) = - 0.12307 \dots

Soluciones generales para

cos (t) = - 0.12307 \dots

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

t = arccos (- 0.12307 \dots) + 2 πn, t = - arccos (- 0.12307 \dots) + 2 πn

t = arccos (- 0.12307 \dots) + 2 πn, t = - arccos (- 0.12307 \dots) + 2 πn

cos (t) = 0.94631 \dots : t = arccos (0.94631 \dots) + 2 πn, t = 2 π - arccos (0.94631 \dots) + 2 πn

cos (t) = 0.94631 \dots

Aplicar propiedades trigonométricas inversas

cos (t) = 0.94631 \dots

Soluciones generales para

cos (t) = 0.94631 \dots

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

t = arccos (0.94631 \dots) + 2 πn, t = 2 π - arccos (0.94631 \dots) + 2 πn

t = arccos (0.94631 \dots) + 2 πn, t = 2 π - arccos (0.94631 \dots) + 2 πn

cos (t) = - 1.07323 \dots :

Sin solución

cos (t) = - 1.07323 \dots

- 1 \leq cos (x) \leq 1

Sin soluci \overset{o}{ˊ} n

Combinar toda las soluciones

t = arccos (- 0.12307 \dots) + 2 πn, t = - arccos (- 0.12307 \dots) + 2 πn, t = arccos (0.94631 \dots) + 2 πn, t = 2 π - arccos (0.94631 \dots) + 2 πn

Mostrar soluciones en forma decimal

t = 1.69418 \dots + 2 πn, t = - 1.69418 \dots + 2 πn, t = 0.32915 \dots + 2 πn, t = 2 π - 0.32915 \dots + 2 πn

Ejemplos populares

sin(B)= 4/5

sin (B) = \frac{4}{5}

cos(2x)=-3cos(x)-2,0<x<= 2pi

cos (2 x) = - 3 cos (x) - 2, 0 < x \leq 2 π

cos(θ)=-7/6

cos (θ) = - \frac{7}{6}

csc(θ)=5,sin(θ)=x

csc (θ) = 5, sin (θ) = x

tan(θ)=0.6

tan (θ) = 0.6