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cos^4(a)+cos^2(a)=1

Solución

cos^{4} (a) + cos^{2} (a) = 1

Solución

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

Grados

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

Pasos de solución

cos^{4} (a) + cos^{2} (a) = 1

Usando el método de sustitución

cos^{4} (a) + cos^{2} (a) = 1

Sea:

cos (a) = u

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

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

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

Mover

1

al lado izquierdo

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

Restar

1

de ambos lados

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

Simplificar

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

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

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

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

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

Aplicar propiedades trigonométricas inversas

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

Soluciones generales para

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

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

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

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

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

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

Aplicar propiedades trigonométricas inversas

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

Soluciones generales para

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

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

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

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

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

cos (a) = \frac{- 1 - 5}{2}

Aplicar propiedades trigonométricas inversas

cos (a) = \frac{- 1 - 5}{2}

Soluciones generales para

cos (a) = \frac{- 1 - 5}{2}

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

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

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

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

cos (a) = - \frac{- 1 - 5}{2}

Aplicar propiedades trigonométricas inversas

cos (a) = - \frac{- 1 - 5}{2}

Soluciones generales para

cos (a) = - \frac{- 1 - 5}{2}

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

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

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

Combinar toda las soluciones

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

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

Mostrar soluciones en forma decimal

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

Ejemplos populares

1-cos(x)=2+cos(x)

1 - cos (x) = 2 + cos (x)

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(5sin(x)+6)/(sin(x))=17

\frac{5 sin ( x ) + 6}{sin ( x )} = 17

tan^2(x)sec(x)-1=0

tan^{2} (x) sec (x) - 1 = 0

sin^2(x)+cos^4(x)=2

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