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Popular >

cos^2(θ)+cos^4(θ)=1

Solución

cos^{2} (θ) + cos^{4} (θ) = 1

Solución

θ = 0.66623 \dots + 2 πn, θ = 2 π - 0.66623 \dots + 2 πn, θ = 2.47535 \dots + 2 πn, θ = - 2.47535 \dots + 2 πn

Grados

θ = 38.17270 \dots^{\circ} + 36 0^{\circ} n, θ = 321.82729 \dots^{\circ} + 36 0^{\circ} n, θ = 141.82729 \dots^{\circ} + 36 0^{\circ} n, θ = - 141.82729 \dots^{\circ} + 36 0^{\circ} n

Pasos de solución

cos^{2} (θ) + cos^{4} (θ) = 1

Usando el método de sustitución

cos^{2} (θ) + cos^{4} (θ) = 1

Sea:

cos (θ) = u

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

u^{2} + u^{4} = 1 : u = \frac{- 1 + 5}{2}, u = - \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}, u = - \frac{- 1 - 5}{2}

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

Mover

1

al lado izquierdo

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

Restar

1

de ambos lados

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

Simplificar

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

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

Escribir en la forma binómica

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

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

Re-escribir la ecuación con

v = u^{2}

v^{2} = u^{4}

v^{2} + v - 1 = 0

Resolver

v^{2} + v - 1 = 0 : v = \frac{- 1 + 5}{2}, v = \frac{- 1 - 5}{2}

v^{2} + v - 1 = 0

Resolver con la fórmula general para ecuaciones de segundo grado:

v^{2} + v - 1 = 0

Formula general para ecuaciones de segundo grado:

Para

a = 1, b = 1, c = - 1

v_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

v_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 1) = 5

1^{2} - 4 \cdot 1 \cdot (- 1)

Aplicar la regla

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 1)

Aplicar la regla

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

Multiplicar los numeros:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

Sumar:

1 + 4 = 5

= 5

v_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 1}

Separar las soluciones

v_{1} = \frac{- 1 + 5}{2 \cdot 1}, v_{2} = \frac{- 1 - 5}{2 \cdot 1}

v = \frac{- 1 + 5}{2 \cdot 1} : \frac{- 1 + 5}{2}

\frac{- 1 + 5}{2 \cdot 1}

Multiplicar los numeros:

2 \cdot 1 = 2

= \frac{- 1 + 5}{2}

v = \frac{- 1 - 5}{2 \cdot 1} : \frac{- 1 - 5}{2}

\frac{- 1 - 5}{2 \cdot 1}

Multiplicar los numeros:

2 \cdot 1 = 2

= \frac{- 1 - 5}{2}

Las soluciones a la ecuación de segundo grado son:

v = \frac{- 1 + 5}{2}, v = \frac{- 1 - 5}{2}

v = \frac{- 1 + 5}{2}, v = \frac{- 1 - 5}{2}

Sustituir hacia atrás la

v = u^{2},

resolver para

u

Resolver

u^{2} = \frac{- 1 + 5}{2} : u = \frac{- 1 + 5}{2}, u = - \frac{- 1 + 5}{2}

u^{2} = \frac{- 1 + 5}{2}

Para

x^{2} = f (a)

las soluciones son

x = f (a), - f (a)

u = \frac{- 1 + 5}{2}, u = - \frac{- 1 + 5}{2}

Resolver

u^{2} = \frac{- 1 - 5}{2} : u = \frac{- 1 - 5}{2}, u = - \frac{- 1 - 5}{2}

u^{2} = \frac{- 1 - 5}{2}

Para

x^{2} = f (a)

las soluciones son

x = f (a), - f (a)

u = \frac{- 1 - 5}{2}, u = - \frac{- 1 - 5}{2}

Las soluciones son

u = \frac{- 1 + 5}{2}, u = - \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}, u = - \frac{- 1 - 5}{2}

Sustituir en la ecuación

u = cos (θ)

cos (θ) = \frac{- 1 + 5}{2}, cos (θ) = - \frac{- 1 + 5}{2}, cos (θ) = \frac{- 1 - 5}{2}, cos (θ) = - \frac{- 1 - 5}{2}

cos (θ) = \frac{- 1 + 5}{2}, cos (θ) = - \frac{- 1 + 5}{2}, cos (θ) = \frac{- 1 - 5}{2}, cos (θ) = - \frac{- 1 - 5}{2}

cos (θ) = \frac{- 1 + 5}{2} : θ = arccos \frac{- 1 + 5}{2} + 2 πn, θ = 2 π - arccos \frac{- 1 + 5}{2} + 2 πn

cos (θ) = \frac{- 1 + 5}{2}

Aplicar propiedades trigonométricas inversas

cos (θ) = \frac{- 1 + 5}{2}

Soluciones generales para

cos (θ) = \frac{- 1 + 5}{2}

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

θ = arccos \frac{- 1 + 5}{2} + 2 πn, θ = 2 π - arccos \frac{- 1 + 5}{2} + 2 πn

θ = arccos \frac{- 1 + 5}{2} + 2 πn, θ = 2 π - arccos \frac{- 1 + 5}{2} + 2 πn

cos (θ) = - \frac{- 1 + 5}{2} : θ = arccos - \frac{- 1 + 5}{2} + 2 πn, θ = - arccos - \frac{- 1 + 5}{2} + 2 πn

cos (θ) = - \frac{- 1 + 5}{2}

Aplicar propiedades trigonométricas inversas

cos (θ) = - \frac{- 1 + 5}{2}

Soluciones generales para

cos (θ) = - \frac{- 1 + 5}{2}

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

θ = arccos - \frac{- 1 + 5}{2} + 2 πn, θ = - arccos - \frac{- 1 + 5}{2} + 2 πn

θ = arccos - \frac{- 1 + 5}{2} + 2 πn, θ = - arccos - \frac{- 1 + 5}{2} + 2 πn

cos (θ) = \frac{- 1 - 5}{2} : θ = arccos \frac{- 1 - 5}{2} + 2 πn, θ = - arccos \frac{- 1 - 5}{2} + 2 πn

cos (θ) = \frac{- 1 - 5}{2}

Aplicar propiedades trigonométricas inversas

cos (θ) = \frac{- 1 - 5}{2}

Soluciones generales para

cos (θ) = \frac{- 1 - 5}{2}

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

θ = arccos \frac{- 1 - 5}{2} + 2 πn, θ = - arccos \frac{- 1 - 5}{2} + 2 πn

θ = arccos \frac{- 1 - 5}{2} + 2 πn, θ = - arccos \frac{- 1 - 5}{2} + 2 πn

cos (θ) = - \frac{- 1 - 5}{2} : θ = arccos - \frac{- 1 - 5}{2} + 2 πn, θ = - arccos - \frac{- 1 - 5}{2} + 2 πn

cos (θ) = - \frac{- 1 - 5}{2}

Aplicar propiedades trigonométricas inversas

cos (θ) = - \frac{- 1 - 5}{2}

Soluciones generales para

cos (θ) = - \frac{- 1 - 5}{2}

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

θ = arccos - \frac{- 1 - 5}{2} + 2 πn, θ = - arccos - \frac{- 1 - 5}{2} + 2 πn

θ = arccos - \frac{- 1 - 5}{2} + 2 πn, θ = - arccos - \frac{- 1 - 5}{2} + 2 πn

Combinar toda las soluciones

θ = arccos \frac{- 1 + 5}{2} + 2 πn, θ = 2 π - arccos \frac{- 1 + 5}{2} + 2 πn, θ = arccos - \frac{- 1 + 5}{2} + 2 πn, θ = - arccos - \frac{- 1 + 5}{2} + 2 πn, θ = arccos \frac{- 1 - 5}{2} + 2 πn, θ = - arccos \frac{- 1 - 5}{2} + 2 πn, θ = arccos - \frac{- 1 - 5}{2} + 2 πn, θ = - arccos - \frac{- 1 - 5}{2} + 2 πn

Siendo que la ecuación esta indefinida para:

arccos \frac{- 1 - 5}{2} + 2 πn, - arccos \frac{- 1 - 5}{2} + 2 πn, arccos - \frac{- 1 - 5}{2} + 2 πn, - arccos - \frac{- 1 - 5}{2} + 2 πn

θ = arccos \frac{- 1 + 5}{2} + 2 πn, θ = 2 π - arccos \frac{- 1 + 5}{2} + 2 πn, θ = arccos - \frac{- 1 + 5}{2} + 2 πn, θ = - arccos - \frac{- 1 + 5}{2} + 2 πn

Mostrar soluciones en forma decimal

θ = 0.66623 \dots + 2 πn, θ = 2 π - 0.66623 \dots + 2 πn, θ = 2.47535 \dots + 2 πn, θ = - 2.47535 \dots + 2 πn

Ejemplos populares

cot(a)=-1/(sqrt(3))

cot (a) = - \frac{1}{3}

2cos(θ)-2sec(θ)-3=0

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

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

sin(θ)= 2/9

sin (θ) = \frac{2}{9}

-2sin^2(θ)+9sin(θ)=8sin(θ)-1

- 2 sin^{2} (θ) + 9 sin (θ) = 8 sin (θ) - 1