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Popular Trigonometría >

5cos^2(x)= 1/(2(1+cos^2(x)))

Solución

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

Solución

x = 1.26330 \dots + 2 πn, x = 2 π - 1.26330 \dots + 2 πn, x = 1.87828 \dots + 2 πn, x = - 1.87828 \dots + 2 πn

Grados

x = 72.38207 \dots^{\circ} + 36 0^{\circ} n, x = 287.61792 \dots^{\circ} + 36 0^{\circ} n, x = 107.61792 \dots^{\circ} + 36 0^{\circ} n, x = - 107.61792 \dots^{\circ} + 36 0^{\circ} n

Pasos de solución

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

Usando el método de sustitución

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

Sea:

cos (x) = u

5 u^{2} = \frac{1}{2 ( 1 + u ^{2} )}

5 u^{2} = \frac{1}{2 ( 1 + u ^{2} )} : u = \frac{- 5 + 35}{10}, u = - \frac{- 5 + 35}{10}, u = i \frac{5 + 35}{10}, u = - i \frac{5 + 35}{10}

5 u^{2} = \frac{1}{2 ( 1 + u ^{2} )}

Multiplicar ambos lados por

2 (1 + u^{2})

5 u^{2} = \frac{1}{2 ( 1 + u ^{2} )}

Multiplicar ambos lados por

2 (1 + u^{2})

5 u^{2} \cdot 2 (1 + u^{2}) = \frac{1}{2 ( 1 + u ^{2} )} \cdot 2 (1 + u^{2})

Simplificar

5 u^{2} \cdot 2 (1 + u^{2}) = \frac{1}{2 ( 1 + u ^{2} )} \cdot 2 (1 + u^{2})

Simplificar

5 u^{2} \cdot 2 (1 + u^{2}) : 10 u^{2} (1 + u^{2})

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

Multiplicar los numeros:

5 \cdot 2 = 10

= 10 u^{2} (u^{2} + 1)

Simplificar

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

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

Multiplicar fracciones:

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

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

Eliminar los terminos comunes:

2

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

Eliminar los terminos comunes:

1 + u^{2}

= 1

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

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

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

Resolver

10 u^{2} (1 + u^{2}) = 1 : u = \frac{- 5 + 35}{10}, u = - \frac{- 5 + 35}{10}, u = i \frac{5 + 35}{10}, u = - i \frac{5 + 35}{10}

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

Desarrollar

10 u^{2} (1 + u^{2}) : 10 u^{2} + 10 u^{4}

10 u^{2} (1 + u^{2})

Poner los parentesis utilizando:

a (b + c) = ab + a c

a = 10 u^{2}, b = 1, c = u^{2}

= 10 u^{2} \cdot 1 + 10 u^{2} u^{2}

= 10 \cdot 1 \cdot u^{2} + 10 u^{2} u^{2}

Simplificar

10 \cdot 1 \cdot u^{2} + 10 u^{2} u^{2} : 10 u^{2} + 10 u^{4}

10 \cdot 1 \cdot u^{2} + 10 u^{2} u^{2}

10 \cdot 1 \cdot u^{2} = 10 u^{2}

10 \cdot 1 \cdot u^{2}

Multiplicar los numeros:

10 \cdot 1 = 10

= 10 u^{2}

10 u^{2} u^{2} = 10 u^{4}

10 u^{2} u^{2}

Aplicar las leyes de los exponentes:

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

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

= 10 u^{2 + 2}

Sumar:

2 + 2 = 4

= 10 u^{4}

= 10 u^{2} + 10 u^{4}

= 10 u^{2} + 10 u^{4}

10 u^{2} + 10 u^{4} = 1

Desplace

1

a la izquierda

10 u^{2} + 10 u^{4} = 1

Restar

1

de ambos lados

10 u^{2} + 10 u^{4} - 1 = 1 - 1

Simplificar

10 u^{2} + 10 u^{4} - 1 = 0

10 u^{2} + 10 u^{4} - 1 = 0

Escribir en la forma binómica

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

10 u^{4} + 10 u^{2} - 1 = 0

Re-escribir la ecuación con

v = u^{2}

v^{2} = u^{4}

10 v^{2} + 10 v - 1 = 0

Resolver

10 v^{2} + 10 v - 1 = 0 : v = \frac{- 5 + 35}{10}, v = - \frac{5 + 35}{10}

10 v^{2} + 10 v - 1 = 0

Resolver con la fórmula general para ecuaciones de segundo grado:

10 v^{2} + 10 v - 1 = 0

Formula general para ecuaciones de segundo grado:

Para

a = 10, b = 10, c = - 1

v_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 10 ( - 1 )}{2 \cdot 10}

v_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 10 ( - 1 )}{2 \cdot 10}

1 0^{2} - 4 \cdot 10 (- 1) = 235

1 0^{2} - 4 \cdot 10 (- 1)

Aplicar la regla

- (- a) = a

= 1 0^{2} + 4 \cdot 10 \cdot 1

Multiplicar los numeros:

4 \cdot 10 \cdot 1 = 40

= 1 0^{2} + 40

1 0^{2} = 100

= 100 + 40

Sumar:

100 + 40 = 140

= 140

Descomposición en factores primos de

140 : 2^{2} \cdot 5 \cdot 7

140

140

divida por

2140 = 70 \cdot 2

= 2 \cdot 70

70

divida por

270 = 35 \cdot 2

= 2 \cdot 2 \cdot 35

35

divida por

535 = 7 \cdot 5

= 2 \cdot 2 \cdot 5 \cdot 7

2, 5, 7

son números primos, por lo tanto, no se pueden factorizar mas

= 2 \cdot 2 \cdot 5 \cdot 7

= 2^{2} \cdot 5 \cdot 7

= 2^{2} \cdot 5 \cdot 7

Aplicar las leyes de los exponentes:

= 2^{2} 5 \cdot 7

Aplicar las leyes de los exponentes:

2^{2} = 2

= 2 5 \cdot 7

Simplificar

= 235

v_{1, 2} = \frac{- 10 \pm 2 35}{2 \cdot 10}

Separar las soluciones

v_{1} = \frac{- 10 + 2 35}{2 \cdot 10}, v_{2} = \frac{- 10 - 2 35}{2 \cdot 10}

v = \frac{- 10 + 2 35}{2 \cdot 10} : \frac{- 5 + 35}{10}

\frac{- 10 + 2 35}{2 \cdot 10}

Multiplicar los numeros:

2 \cdot 10 = 20

= \frac{- 10 + 2 35}{20}

Factorizar

- 10 + 235 : 2 (- 5 + 35)

- 10 + 235

Reescribir como

= - 2 \cdot 5 + 235

Factorizar el termino común

2

= 2 (- 5 + 35)

= \frac{2 ( - 5 + 35 )}{20}

Eliminar los terminos comunes:

2

= \frac{- 5 + 35}{10}

v = \frac{- 10 - 2 35}{2 \cdot 10} : - \frac{5 + 35}{10}

\frac{- 10 - 2 35}{2 \cdot 10}

Multiplicar los numeros:

2 \cdot 10 = 20

= \frac{- 10 - 2 35}{20}

Factorizar

- 10 - 235 : - 2 (5 + 35)

- 10 - 235

Reescribir como

= - 2 \cdot 5 - 235

Factorizar el termino común

2

= - 2 (5 + 35)

= - \frac{2 ( 5 + 35 )}{20}

Eliminar los terminos comunes:

2

= - \frac{5 + 35}{10}

Las soluciones a la ecuación de segundo grado son:

v = \frac{- 5 + 35}{10}, v = - \frac{5 + 35}{10}

v = \frac{- 5 + 35}{10}, v = - \frac{5 + 35}{10}

Sustituir hacia atrás la

v = u^{2},

resolver para

u

Resolver

u^{2} = \frac{- 5 + 35}{10} : u = \frac{- 5 + 35}{10}, u = - \frac{- 5 + 35}{10}

u^{2} = \frac{- 5 + 35}{10}

Para

x^{2} = f (a)

las soluciones son

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

u = \frac{- 5 + 35}{10}, u = - \frac{- 5 + 35}{10}

Resolver

u^{2} = - \frac{5 + 35}{10} : u = i \frac{5 + 35}{10}, u = - i \frac{5 + 35}{10}

u^{2} = - \frac{5 + 35}{10}

Para

x^{2} = f (a)

las soluciones son

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

u = - \frac{5 + 35}{10}, u = - - \frac{5 + 35}{10}

Simplificar

- \frac{5 + 35}{10} : i \frac{5 + 35}{10}

- \frac{5 + 35}{10}

Aplicar las leyes de los exponentes:

- a = - 1 a

- \frac{5 + 35}{10} = - 1 \frac{5 + 35}{10}

= - 1 \frac{5 + 35}{10}

Aplicar las propiedades de los numeros imaginarios:

- 1 = i

= i \frac{5 + 35}{10}

Simplificar

- - \frac{5 + 35}{10} : - i \frac{5 + 35}{10}

- - \frac{5 + 35}{10}

Simplificar

- \frac{5 + 35}{10} : i \frac{5 + 35}{10}

- \frac{5 + 35}{10}

Aplicar las leyes de los exponentes:

- a = - 1 a

- \frac{5 + 35}{10} = - 1 \frac{5 + 35}{10}

= - 1 \frac{5 + 35}{10}

Aplicar las propiedades de los numeros imaginarios:

- 1 = i

= i \frac{5 + 35}{10}

= - i \frac{5 + 35}{10}

u = i \frac{5 + 35}{10}, u = - i \frac{5 + 35}{10}

Las soluciones son

u = \frac{- 5 + 35}{10}, u = - \frac{- 5 + 35}{10}, u = i \frac{5 + 35}{10}, u = - i \frac{5 + 35}{10}

u = \frac{- 5 + 35}{10}, u = - \frac{- 5 + 35}{10}, u = i \frac{5 + 35}{10}, u = - i \frac{5 + 35}{10}

Sustituir en la ecuación

u = cos (x)

cos (x) = \frac{- 5 + 35}{10}, cos (x) = - \frac{- 5 + 35}{10}, cos (x) = i \frac{5 + 35}{10}, cos (x) = - i \frac{5 + 35}{10}

cos (x) = \frac{- 5 + 35}{10}, cos (x) = - \frac{- 5 + 35}{10}, cos (x) = i \frac{5 + 35}{10}, cos (x) = - i \frac{5 + 35}{10}

cos (x) = \frac{- 5 + 35}{10} : x = arccos \frac{- 5 + 35}{10} + 2 πn, x = 2 π - arccos \frac{- 5 + 35}{10} + 2 πn

cos (x) = \frac{- 5 + 35}{10}

Aplicar propiedades trigonométricas inversas

cos (x) = \frac{- 5 + 35}{10}

Soluciones generales para

cos (x) = \frac{- 5 + 35}{10}

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

x = arccos \frac{- 5 + 35}{10} + 2 πn, x = 2 π - arccos \frac{- 5 + 35}{10} + 2 πn

x = arccos \frac{- 5 + 35}{10} + 2 πn, x = 2 π - arccos \frac{- 5 + 35}{10} + 2 πn

cos (x) = - \frac{- 5 + 35}{10} : x = arccos - \frac{- 5 + 35}{10} + 2 πn, x = - arccos - \frac{- 5 + 35}{10} + 2 πn

cos (x) = - \frac{- 5 + 35}{10}

Aplicar propiedades trigonométricas inversas

cos (x) = - \frac{- 5 + 35}{10}

Soluciones generales para

cos (x) = - \frac{- 5 + 35}{10}

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

x = arccos - \frac{- 5 + 35}{10} + 2 πn, x = - arccos - \frac{- 5 + 35}{10} + 2 πn

x = arccos - \frac{- 5 + 35}{10} + 2 πn, x = - arccos - \frac{- 5 + 35}{10} + 2 πn

cos (x) = i \frac{5 + 35}{10} :

Sin solución

cos (x) = i \frac{5 + 35}{10}

Sin soluci \overset{o}{ˊ} n

cos (x) = - i \frac{5 + 35}{10} :

Sin solución

cos (x) = - i \frac{5 + 35}{10}

Sin soluci \overset{o}{ˊ} n

Combinar toda las soluciones

x = arccos \frac{- 5 + 35}{10} + 2 πn, x = 2 π - arccos \frac{- 5 + 35}{10} + 2 πn, x = arccos - \frac{- 5 + 35}{10} + 2 πn, x = - arccos - \frac{- 5 + 35}{10} + 2 πn

Mostrar soluciones en forma decimal

x = 1.26330 \dots + 2 πn, x = 2 π - 1.26330 \dots + 2 πn, x = 1.87828 \dots + 2 πn, x = - 1.87828 \dots + 2 πn

Gráfica

Ejemplos populares

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