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Popular >

1=sqrt(4cos(2θ))

Solución

1 = 4 cos (2 θ)

Solución

θ = \frac{1.31811 \dots}{2} + πn, θ = π - \frac{1.31811 \dots}{2} + πn

Grados

θ = 37.76124 \dots^{\circ} + 18 0^{\circ} n, θ = 142.23875 \dots^{\circ} + 18 0^{\circ} n

Pasos de solución

1 = 4 cos (2 θ)

Intercambiar lados

4 cos (2 θ) = 1

Elevar al cuadrado ambos lados:

4 cos (2 θ) = 1

4 cos (2 θ) = 1

(4 cos (2 θ))^{2} = 1^{2}

Desarrollar

(4 cos (2 θ))^{2} : 4 cos (2 θ)

(4 cos (2 θ))^{2}

Aplicar las leyes de los exponentes:

a = a^{\frac{1}{2}}

= ((4 cos (2 θ))^{\frac{1}{2}})^{2}

Aplicar las leyes de los exponentes:

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

= (4 cos (2 θ))^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiplicar fracciones:

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

= \frac{1 \cdot 2}{2}

Eliminar los terminos comunes:

2

= 1

= 4 cos (2 θ)

Desarrollar

1^{2} : 1

1^{2}

Aplicar la regla

1^{a} = 1

= 1

4 cos (2 θ) = 1

4 cos (2 θ) = 1

Resolver

4 cos (2 θ) = 1 : cos (2 θ) = \frac{1}{4}

4 cos (2 θ) = 1

Dividir ambos lados entre

4

4 cos (2 θ) = 1

Dividir ambos lados entre

4

\frac{4 cos ( 2 θ )}{4} = \frac{1}{4}

Simplificar

cos (2 θ) = \frac{1}{4}

cos (2 θ) = \frac{1}{4}

cos (2 θ) = \frac{1}{4}

Verificar las soluciones:

cos (2 θ) = \frac{1}{4}

Verdadero

Verificar las soluciones sustituyéndolas en

4 cos (2 θ) = 1

Quitar las que no concuerden con la ecuación.

Sustituir

cos (2 θ) = \frac{1}{4} :

Verdadero

4 (\frac{1}{4}) = 1

4 (\frac{1}{4}) = 1

4 (\frac{1}{4})

Quitar los parentesis:

(a) = a

= 4 \cdot \frac{1}{4}

Multiplicar

4 \cdot \frac{1}{4} : 1

4 \cdot \frac{1}{4}

Multiplicar fracciones:

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

= \frac{1 \cdot 4}{4}

Eliminar los terminos comunes:

4

= 1

= 1

Aplicar la regla

1 = 1

= 1

1 = 1

Verdadero

La solución es

cos (2 θ) = \frac{1}{4}

Aplicar propiedades trigonométricas inversas

cos (2 θ) = \frac{1}{4}

Soluciones generales para

cos (2 θ) = \frac{1}{4}

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

2 θ = arccos (\frac{1}{4}) + 2 πn, 2 θ = 2 π - arccos (\frac{1}{4}) + 2 πn

2 θ = arccos (\frac{1}{4}) + 2 πn, 2 θ = 2 π - arccos (\frac{1}{4}) + 2 πn

Resolver

2 θ = arccos (\frac{1}{4}) + 2 πn : θ = \frac{arccos ( \frac{1}{4} )}{2} + πn

2 θ = arccos (\frac{1}{4}) + 2 πn

Dividir ambos lados entre

2

2 θ = arccos (\frac{1}{4}) + 2 πn

Dividir ambos lados entre

2

\frac{2 θ}{2} = \frac{arccos ( \frac{1}{4} )}{2} + \frac{2 πn}{2}

Simplificar

θ = \frac{arccos ( \frac{1}{4} )}{2} + πn

θ = \frac{arccos ( \frac{1}{4} )}{2} + πn

Resolver

2 θ = 2 π - arccos (\frac{1}{4}) + 2 πn : θ = π - \frac{arccos ( \frac{1}{4} )}{2} + πn

2 θ = 2 π - arccos (\frac{1}{4}) + 2 πn

Dividir ambos lados entre

2

2 θ = 2 π - arccos (\frac{1}{4}) + 2 πn

Dividir ambos lados entre

2

\frac{2 θ}{2} = \frac{2 π}{2} - \frac{arccos ( \frac{1}{4} )}{2} + \frac{2 πn}{2}

Simplificar

θ = π - \frac{arccos ( \frac{1}{4} )}{2} + πn

θ = π - \frac{arccos ( \frac{1}{4} )}{2} + πn

θ = \frac{arccos ( \frac{1}{4} )}{2} + πn, θ = π - \frac{arccos ( \frac{1}{4} )}{2} + πn

Mostrar soluciones en forma decimal

θ = \frac{1.31811 \dots}{2} + πn, θ = π - \frac{1.31811 \dots}{2} + πn

Ejemplos populares

209cos(47)=150cos(θ)

209 cos (4 7^{\circ}) = 150 cos (θ)

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7 - 5 cos (x) = 10, 7 - 5 cos (x) = 10

cos(4x)=cos(x)

cos (4 x) = cos (x)

-4cos(2θ)-19=-26cos(θ),0<= θ<360

- 4 cos (2 θ) - 19 = - 26 cos (θ), 0 \leq θ < 36 0^{\circ}

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