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Popular Trigonometría >

n=(2tan^3(2θ)-1)^{1/2}

Solución

n = (2 tan^{3} (2 θ) - 1)^{\frac{1}{2}}

Solución

θ = \frac{arctan ( 3 \frac{n ^{2} + 1}{2} )}{2} + \frac{πk}{2}

Pasos de solución

n = (2 tan^{3} (2 θ) - 1)^{\frac{1}{2}}

Intercambiar lados

(2 tan^{3} (2 θ) - 1)^{\frac{1}{2}} = n

Elevar al cuadrado ambos lados:

2 tan^{3} (2 θ) - 1 = n^{2}

(2 tan^{3} (2 θ) - 1)^{\frac{1}{2}} = n

((2 tan^{3} (2 θ) - 1)^{\frac{1}{2}})^{2} = n^{2}

Desarrollar

((2 tan^{3} (2 θ) - 1)^{\frac{1}{2}})^{2} : 2 tan^{3} (2 θ) - 1

((2 tan^{3} (2 θ) - 1)^{\frac{1}{2}})^{2}

Aplicar las leyes de los exponentes:

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

= (2 tan^{3} (2 θ) - 1)^{\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

= 2 tan^{3} (2 θ) - 1

2 tan^{3} (2 θ) - 1 = n^{2}

2 tan^{3} (2 θ) - 1 = n^{2}

Resolver

2 tan^{3} (2 θ) - 1 = n^{2} : tan (2 θ) = 3 \frac{n ^{2} + 1}{2}

2 tan^{3} (2 θ) - 1 = n^{2}

Desplace

1

a la derecha

2 tan^{3} (2 θ) - 1 = n^{2}

Sumar

1

a ambos lados

2 tan^{3} (2 θ) - 1 + 1 = n^{2} + 1

Simplificar

2 tan^{3} (2 θ) = n^{2} + 1

2 tan^{3} (2 θ) = n^{2} + 1

Dividir ambos lados entre

2

2 tan^{3} (2 θ) = n^{2} + 1

Dividir ambos lados entre

2

\frac{2 tan ^{3} ( 2 θ )}{2} = \frac{n ^{2}}{2} + \frac{1}{2}

Simplificar

\frac{2 tan ^{3} ( 2 θ )}{2} = \frac{n ^{2}}{2} + \frac{1}{2}

Simplificar

\frac{2 tan ^{3} ( 2 θ )}{2} : tan^{3} (2 θ)

\frac{2 tan ^{3} ( 2 θ )}{2}

Dividir:

\frac{2}{2} = 1

= tan^{3} (2 θ)

Simplificar

\frac{n ^{2}}{2} + \frac{1}{2} : \frac{n ^{2} + 1}{2}

\frac{n ^{2}}{2} + \frac{1}{2}

Aplicar la regla

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

= \frac{n ^{2} + 1}{2}

tan^{3} (2 θ) = \frac{n ^{2} + 1}{2}

tan^{3} (2 θ) = \frac{n ^{2} + 1}{2}

tan^{3} (2 θ) = \frac{n ^{2} + 1}{2}

Para

x^{n} = f (a)

, n es impar, la solución es

x = n f (a)

tan (2 θ) = 3 \frac{n ^{2} + 1}{2}

tan (2 θ) = 3 \frac{n ^{2} + 1}{2}

Verificar las soluciones:

tan (2 θ) = 3 \frac{n ^{2} + 1}{2} {n \geq 0}

Verificar las soluciones sustituyéndolas en

(2 tan^{3} (2 θ) - 1)^{\frac{1}{2}} = n

Quitar las que no concuerden con la ecuación.

Inserir

tan (2 θ) = 3 \frac{n ^{2} + 1}{2} : 2 (3 \frac{n ^{2} + 1}{2})^{3} - 1^{\frac{1}{2}} = n \Rightarrow n \geq 0

2 (3 \frac{n ^{2} + 1}{2})^{3} - 1^{\frac{1}{2}} = n

Elevar al cuadrado ambos lados:

n^{2} = n^{2}

2 (3 \frac{n ^{2} + 1}{2})^{3} - 1^{\frac{1}{2}} = n

2 (3 \frac{n ^{2} + 1}{2})^{3} - 1^{\frac{1}{2}}^{2} = n^{2}

Desarrollar

2 (3 \frac{n ^{2} + 1}{2})^{3} - 1^{\frac{1}{2}}^{2} : n^{2}

2 (3 \frac{n ^{2} + 1}{2})^{3} - 1^{\frac{1}{2}}^{2}

Aplicar las leyes de los exponentes:

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

= 2 (3 \frac{n ^{2} + 1}{2})^{3} - 1^{\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

= 2 (3 \frac{n ^{2} + 1}{2})^{3} - 1

Desarrollar

2 (3 \frac{n ^{2} + 1}{2})^{3} - 1 : n^{2}

2 (3 \frac{n ^{2} + 1}{2})^{3} - 1

2 (3 \frac{n ^{2} + 1}{2})^{3} = n^{2} + 1

2 (3 \frac{n ^{2} + 1}{2})^{3}

(3 \frac{n ^{2} + 1}{2})^{3} = \frac{n ^{2} + 1}{2}

(3 \frac{n ^{2} + 1}{2})^{3}

Aplicar las leyes de los exponentes:

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

= (\frac{n ^{2} + 1}{2})^{\frac{1}{3} \cdot 3}

\frac{1}{3} \cdot 3 = 1

\frac{1}{3} \cdot 3

Multiplicar fracciones:

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

= \frac{1 \cdot 3}{3}

Eliminar los terminos comunes:

3

= 1

= \frac{n ^{2} + 1}{2}

= 2 \cdot \frac{n ^{2} + 1}{2}

Multiplicar fracciones:

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

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

Eliminar los terminos comunes:

2

= n^{2} + 1

= n^{2} + 1 - 1

1 - 1 = 0

= n^{2}

= n^{2}

n^{2} = n^{2}

n^{2} = n^{2}

Los lados son iguales

Verdadero para todo n

Verificar las soluciones:

n < 0

Falso

, n = 0

Verdadero

, n > 0

Verdadero

2 (3 \frac{n ^{2} + 1}{2})^{3} - 1^{\frac{1}{2}} = n

Combinar intervalo de dominio con el de solución:

Verdadero para todo n

Encontrar los intervalos de la función:

n < 0, n = 0, n > 0

2 (3 \frac{n ^{2} + 1}{2})^{3} - 1^{\frac{1}{2}} = n

Encontrar las raíces pares con argumento cero:

Resolver

2 3 \frac{n ^{2} + 1}{2}^{3} - 1 = 0 : n = 0

2 (3 \frac{n ^{2} + 1}{2})^{3} - 1 = 0

Factorizar

2 (3 \frac{n ^{2} + 1}{2})^{3} - 1 : (2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} - 1) (2^{\frac{2}{3}} (\frac{n ^{2} + 1}{2})^{\frac{2}{3}} + 2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} + 1)

2 (3 \frac{n ^{2} + 1}{2})^{3} - 1

Reescribir

2 (3 \frac{n ^{2} + 1}{2})^{3} - 1

como

(2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2})^{3} - 1^{3}

2 (3 \frac{n ^{2} + 1}{2})^{3} - 1

Reescribir

2

como

(2^{\frac{1}{3}})^{3}

= (2^{\frac{1}{3}})^{3} (3 \frac{n ^{2} + 1}{2})^{3} - 1

Reescribir

1

como

1^{3}

= (2^{\frac{1}{3}})^{3} (3 \frac{n ^{2} + 1}{2})^{3} - 1^{3}

Aplicar las leyes de los exponentes:

a^{m} b^{m} = (ab)^{m}

(2^{\frac{1}{3}})^{3} (3 \frac{n ^{2} + 1}{2})^{3} = (2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2})^{3}

= (2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2})^{3} - 1^{3}

= (2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2})^{3} - 1^{3}

Aplicar la siguiente regla de productos notables (Diferencia de cubos):

x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})

(2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2})^{3} - 1^{3} = (2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} - 1) (2^{\frac{1}{3}})^{2} (3 \frac{n ^{2} + 1}{2})^{2} + 2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} + 1

= (2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} - 1) (2^{\frac{1}{3}})^{2} (3 \frac{n ^{2} + 1}{2})^{2} + 2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} + 1

Simplificar

= (2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} - 1) 2^{\frac{2}{3}} (3 \frac{n ^{2} + 1}{2})^{2} + 2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} + 1

(3 \frac{n ^{2} + 1}{2})^{2} = \frac{n ^{2} + 1}{2}^{\frac{2}{3}}

(3 \frac{n ^{2} + 1}{2})^{2}

Aplicar las leyes de los exponentes:

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

= (\frac{n ^{2} + 1}{2})^{\frac{1}{3} \cdot 2}

\frac{1}{3} \cdot 2 = \frac{2}{3}

\frac{1}{3} \cdot 2

Multiplicar fracciones:

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

= \frac{1 \cdot 2}{3}

Multiplicar los numeros:

1 \cdot 2 = 2

= \frac{2}{3}

= (\frac{n ^{2} + 1}{2})^{\frac{2}{3}}

= (2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} - 1) (2^{\frac{2}{3}} (\frac{n ^{2} + 1}{2})^{\frac{2}{3}} + 2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} + 1)

(2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} - 1) (2^{\frac{2}{3}} (\frac{n ^{2} + 1}{2})^{\frac{2}{3}} + 2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} + 1) = 0

Usando la propiedad del factor cero:

ab = 0

entonces

a = 0

b = 0

2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} - 1 = 0 or 2^{\frac{2}{3}} (\frac{n ^{2} + 1}{2})^{\frac{2}{3}} + 2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} + 1 = 0

Resolver

2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} - 1 = 0 : n = 0

2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} - 1 = 0

Desplace

1

a la derecha

2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} - 1 = 0

Sumar

1

a ambos lados

2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} - 1 + 1 = 0 + 1

Simplificar

2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} = 1

2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} = 1

Dividir ambos lados entre

2^{\frac{1}{3}}

2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} = 1

Dividir ambos lados entre

2^{\frac{1}{3}}

\frac{2 ^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2}}{2 ^{\frac{1}{3}}} = \frac{1}{2 ^{\frac{1}{3}}}

Simplificar

3 \frac{n ^{2} + 1}{2} = \frac{1}{2 ^{\frac{1}{3}}}

3 \frac{n ^{2} + 1}{2} = \frac{1}{2 ^{\frac{1}{3}}}

Elevar ambos lados de la ecuación a la potencia

3 : \frac{n ^{2} + 1}{2} = \frac{1}{2}

3 \frac{n ^{2} + 1}{2} = \frac{1}{2 ^{\frac{1}{3}}}

(3 \frac{n ^{2} + 1}{2})^{3} = (\frac{1}{2 ^{\frac{1}{3}}})^{3}

Desarrollar

(3 \frac{n ^{2} + 1}{2})^{3} : \frac{n ^{2} + 1}{2}

(3 \frac{n ^{2} + 1}{2})^{3}

Aplicar las leyes de los exponentes:

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

= (\frac{n ^{2} + 1}{2})^{\frac{1}{3} \cdot 3}

\frac{1}{3} \cdot 3 = 1

\frac{1}{3} \cdot 3

Multiplicar fracciones:

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

= \frac{1 \cdot 3}{3}

Eliminar los terminos comunes:

3

= 1

= \frac{n ^{2} + 1}{2}

Desarrollar

(\frac{1}{2 ^{\frac{1}{3}}})^{3} : \frac{1}{2}

(\frac{1}{2 ^{\frac{1}{3}}})^{3}

Aplicar las leyes de los exponentes:

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

= \frac{1 ^{3}}{( 2 ^{\frac{1}{3}} ) ^{3}}

(2^{\frac{1}{3}})^{3} : 2

Aplicar las leyes de los exponentes:

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

= 2^{\frac{1}{3} \cdot 3}

\frac{1}{3} \cdot 3 = 1

\frac{1}{3} \cdot 3

Multiplicar fracciones:

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

= \frac{1 \cdot 3}{3}

Eliminar los terminos comunes:

3

= 1

= 2

= \frac{1 ^{3}}{2}

Aplicar la regla

1^{a} = 1

1^{3} = 1

= \frac{1}{2}

\frac{n ^{2} + 1}{2} = \frac{1}{2}

\frac{n ^{2} + 1}{2} = \frac{1}{2}

Resolver

\frac{n ^{2} + 1}{2} = \frac{1}{2} : n = 0

\frac{n ^{2} + 1}{2} = \frac{1}{2}

Multiplicar ambos lados por

2

\frac{n ^{2} + 1}{2} = \frac{1}{2}

Multiplicar ambos lados por

2

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

Simplificar

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

Simplificar

\frac{2 ( n ^{2} + 1 )}{2} : n^{2} + 1

\frac{2 ( n ^{2} + 1 )}{2}

Dividir:

\frac{2}{2} = 1

= n^{2} + 1

Simplificar

\frac{1 \cdot 2}{2} : 1

\frac{1 \cdot 2}{2}

Multiplicar los numeros:

1 \cdot 2 = 2

= \frac{2}{2}

Aplicar la regla

\frac{a}{a} = 1

= 1

n^{2} + 1 = 1

n^{2} + 1 = 1

n^{2} + 1 = 1

Desplace

1

a la derecha

n^{2} + 1 = 1

Restar

1

de ambos lados

n^{2} + 1 - 1 = 1 - 1

Simplificar

n^{2} = 0

n^{2} = 0

Aplicar la regla

x^{n} = 0 \Rightarrow x = 0

n = 0

n = 0

Verificar las soluciones:

n = 0

Verdadero

Verificar las soluciones sustituyéndolas en

2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} - 1 = 0

Quitar las que no concuerden con la ecuación.

Sustituir

n = 0 :

Verdadero

2^{\frac{1}{3}} 3 \frac{0 ^{2} + 1}{2} - 1 = 0

2^{\frac{1}{3}} 3 \frac{0 ^{2} + 1}{2} - 1 = 0

2^{\frac{1}{3}} 3 \frac{0 ^{2} + 1}{2} - 1

Aplicar la regla

0^{a} = 0

0^{2} = 0

= 2^{\frac{1}{3}} 3 \frac{0 + 1}{2} - 1

2^{\frac{1}{3}} 3 \frac{0 + 1}{2} = 1

2^{\frac{1}{3}} 3 \frac{0 + 1}{2}

Sumar:

0 + 1 = 1

= 2^{\frac{1}{3}} 3 \frac{1}{2}

Aplicar las leyes de los exponentes:

a^{m} b^{m} = (ab)^{m}

2^{\frac{1}{3}} 3 \frac{1}{2} = (2 \cdot \frac{1}{2})^{\frac{1}{3}}

= (2 \cdot \frac{1}{2})^{\frac{1}{3}}

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

2 \cdot \frac{1}{2}

Multiplicar fracciones:

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

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

Eliminar los terminos comunes:

2

= 1

= 1^{\frac{1}{3}}

Aplicar la regla

1^{a} = 1

= 1

= 1 - 1

Restar:

1 - 1 = 0

= 0

0 = 0

Verdadero

La solución es

n = 0

Resolver

2^{\frac{2}{3}} (\frac{n ^{2} + 1}{2})^{\frac{2}{3}} + 2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} + 1 = 0 :

Sin solución para

n \in R

2^{\frac{2}{3}} (\frac{n ^{2} + 1}{2})^{\frac{2}{3}} + 2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} + 1 = 0

Usar la siguiente propiedad de los exponentes

: a^{\frac{n}{m}} = (m a)^{n}

(\frac{n ^{2} + 1}{2})^{\frac{2}{3}} = (3 \frac{n ^{2} + 1}{2})^{2}

2^{\frac{2}{3}} (3 \frac{n ^{2} + 1}{2})^{2} + 2^{\frac{1}{3}} 3 \frac{n ^{2} + 1}{2} + 1 = 0

Re escribir la ecuación con

3 \frac{n ^{2} + 1}{2} = u

2^{\frac{2}{3}} u^{2} + 2^{\frac{1}{3}} u + 1 = 0

Resolver

2^{\frac{2}{3}} u^{2} + 2^{\frac{1}{3}} u + 1 = 0 :

Sin solución para

u \in R

2^{\frac{2}{3}} u^{2} + 2^{\frac{1}{3}} u + 1 = 0

Discriminante

2^{\frac{2}{3}} u^{2} + 2^{\frac{1}{3}} u + 1 = 0 : - 3 \cdot 2^{\frac{2}{3}}

2^{\frac{2}{3}} u^{2} + 2^{\frac{1}{3}} u + 1 = 0

Para una ecuación cuadrática de la forma

a x^{2} + b x + c = 0

el discriminante es

b^{2} - 4 a c

Para

a = 2^{\frac{2}{3}}, b = 2^{\frac{1}{3}}, c = 1 : (2^{\frac{1}{3}})^{2} - 4 \cdot 2^{\frac{2}{3}} \cdot 1

(2^{\frac{1}{3}})^{2} - 4 \cdot 2^{\frac{2}{3}} \cdot 1

Desarrollar

(2^{\frac{1}{3}})^{2} - 4 \cdot 2^{\frac{2}{3}} \cdot 1 : - 3 \cdot 2^{\frac{2}{3}}

(2^{\frac{1}{3}})^{2} - 4 \cdot 2^{\frac{2}{3}} \cdot 1

(2^{\frac{1}{3}})^{2} = 2^{\frac{2}{3}}

(2^{\frac{1}{3}})^{2}

Aplicar las leyes de los exponentes:

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

= 2^{\frac{1}{3} \cdot 2}

\frac{1}{3} \cdot 2 = \frac{2}{3}

\frac{1}{3} \cdot 2

Multiplicar fracciones:

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

= \frac{1 \cdot 2}{3}

Multiplicar los numeros:

1 \cdot 2 = 2

= \frac{2}{3}

= 2^{\frac{2}{3}}

4 \cdot 2^{\frac{2}{3}} \cdot 1 = 4 \cdot 2^{\frac{2}{3}}

4 \cdot 2^{\frac{2}{3}} \cdot 1

Multiplicar los numeros:

4 \cdot 1 = 4

= 4 \cdot 2^{\frac{2}{3}}

= 2^{\frac{2}{3}} - 4 \cdot 2^{\frac{2}{3}}

Sumar elementos similares:

2^{\frac{2}{3}} - 4 \cdot 2^{\frac{2}{3}} = - 3 \cdot 2^{\frac{2}{3}}

= - 3 \cdot 2^{\frac{2}{3}}

- 3 \cdot 2^{\frac{2}{3}}

El discriminante no puede ser negativo para

u \in R

La solución es

Sin soluci \overset{o}{ˊ} n para u \in R

Sin soluci \overset{o}{ˊ} n para n \in R

n = 0

Verificar las soluciones:

n = 0

Verdadero

Verificar las soluciones sustituyéndolas en

2 3 \frac{n ^{2} + 1}{2}^{3} - 1 = 0

Quitar las que no concuerden con la ecuación.

Sustituir

n = 0 :

Verdadero

2 (3 \frac{0 ^{2} + 1}{2})^{3} - 1 = 0

2 (3 \frac{0 ^{2} + 1}{2})^{3} - 1 = 0

2 (3 \frac{0 ^{2} + 1}{2})^{3} - 1

Aplicar la regla

0^{a} = 0

0^{2} = 0

= 2 (3 \frac{0 + 1}{2})^{3} - 1

2 (3 \frac{0 + 1}{2})^{3} = 1

2 (3 \frac{0 + 1}{2})^{3}

(3 \frac{0 + 1}{2})^{3} = \frac{1}{2}

(3 \frac{0 + 1}{2})^{3}

Aplicar las leyes de los exponentes:

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

= (\frac{0 + 1}{2})^{\frac{1}{3} \cdot 3}

\frac{1}{3} \cdot 3 = 1

\frac{1}{3} \cdot 3

Multiplicar fracciones:

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

= \frac{1 \cdot 3}{3}

Eliminar los terminos comunes:

3

= 1

= \frac{0 + 1}{2}

Sumar:

0 + 1 = 1

= \frac{1}{2}

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

Multiplicar fracciones:

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

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

Eliminar los terminos comunes:

2

= 1

= 1 - 1

Restar:

1 - 1 = 0

= 0

0 = 0

Verdadero

La solución es

n = 0

n = 0

Los intervalos estan definidos alrededor de los ceros:

n < 0, n = 0, n > 0

Combinar intervalos con el dominio

n < 0, n = 0, n > 0

Verificar las soluciones sustituyéndolas en

(2 3 \frac{n ^{2} + 1}{2}^{3} - 1)^{\frac{1}{2}} = n

Quitar las que no concuerden con la ecuación.

Sustituir

n < 0 : (2 3 \frac{n ^{2} + 1}{2}^{3} - 1)^{\frac{1}{2}} \neq = n \Rightarrow

Falso

La solución es

n \geq 0

La solución es

tan (2 θ) = 3 \frac{n ^{2} + 1}{2} {n \geq 0}

Aplicar propiedades trigonométricas inversas

tan (2 θ) = 3 \frac{n ^{2} + 1}{2}

Soluciones generales para

tan (2 θ) = 3 \frac{n ^{2} + 1}{2}

tan (x) = a \Rightarrow x = arctan (a) + πk

2 θ = arctan (3 \frac{n ^{2} + 1}{2}) + πk

2 θ = arctan (3 \frac{n ^{2} + 1}{2}) + πk

Resolver

2 θ = arctan (3 \frac{n ^{2} + 1}{2}) + πk : θ = \frac{arctan ( 3 \frac{n ^{2} + 1}{2} )}{2} + \frac{πk}{2}

2 θ = arctan (3 \frac{n ^{2} + 1}{2}) + πk

Dividir ambos lados entre

2

2 θ = arctan (3 \frac{n ^{2} + 1}{2}) + πk

Dividir ambos lados entre

2

\frac{2 θ}{2} = \frac{arctan ( 3 \frac{n ^{2} + 1}{2} )}{2} + \frac{πk}{2}

Simplificar

θ = \frac{arctan ( 3 \frac{n ^{2} + 1}{2} )}{2} + \frac{πk}{2}

θ = \frac{arctan ( 3 \frac{n ^{2} + 1}{2} )}{2} + \frac{πk}{2}

θ = \frac{arctan ( 3 \frac{n ^{2} + 1}{2} )}{2} + \frac{πk}{2}

Gráfica

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sin (5 x + 8) = cos (9 x - 16)

cos(x)=(sqrt(125))/(sqrt(174))

cos (x^{\circ}) = \frac{125}{174}