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(sin(x))^{(sin(x))}= 1/(sqrt(2))

Solución

(sin (x))^{(s i n (x))} = \frac{1}{2}

Solución

x = 0.25268 \dots + 2 πn, x = π - 0.25268 \dots + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Grados

x = 14.47751 \dots^{\circ} + 36 0^{\circ} n, x = 165.52248 \dots^{\circ} + 36 0^{\circ} n, x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

Pasos de solución

(sin (x))^{(s i n (x))} = \frac{1}{2}

Restar

\frac{1}{2}

de ambos lados

sin^{s i n (x)} (x) - \frac{1}{2} = 0

Simplificar

sin^{s i n (x)} (x) - \frac{1}{2} : \frac{2 sin ^{s i n (x)} ( x ) - 1}{2}

sin^{s i n (x)} (x) - \frac{1}{2}

Convertir a fracción:

sin^{s i n (x)} (x) = \frac{sin ^{s i n (x)} ( x ) 2}{2}

= \frac{sin ^{s i n (x)} ( x ) 2}{2} - \frac{1}{2}

Ya que los denominadores son iguales, combinar las fracciones:

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

= \frac{sin ^{s i n (x)} ( x ) 2 - 1}{2}

\frac{2 sin ^{s i n (x)} ( x ) - 1}{2} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

2 sin^{s i n (x)} (x) - 1 = 0

Re-escribir usando identidades trigonométricas

- 1 + sin^{s i n (x)} (x) 2

Utilizar la identidad trigonométrica básica:

sin (x) = \frac{1}{csc ( x )}

= - 1 + (\frac{1}{csc ( x )})^{\frac{1}{c s c ( x )}} 2

- 1 + (\frac{1}{csc ( x )})^{\frac{1}{c s c ( x )}} 2 = 0

Usando el método de sustitución

- 1 + (\frac{1}{csc ( x )})^{\frac{1}{c s c ( x )}} 2 = 0

Sea:

csc (x) = u

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

- 1 + (\frac{1}{u})^{\frac{1}{u}} 2 = 0 : u = 4, u = 2

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

Aplicar las leyes de los exponentes

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

Aplicar las leyes de los exponentes:

f (x)^{g (x)} = e^{g (x) l n (f (x))}

(\frac{1}{u})^{\frac{1}{u}} = e^{\frac{1}{u} l n (\frac{1}{u})}

- 1 + e^{\frac{1}{u} l n (\frac{1}{u})} 2 = 0

- 1 + e^{\frac{1}{u} l n (\frac{1}{u})} 2 = 0

Sumar

1

a ambos lados

- 1 + e^{\frac{1}{u} l n (\frac{1}{u})} 2 + 1 = 0 + 1

Simplificar

2 e^{\frac{1}{u} l n (\frac{1}{u})} = 1

Dividir ambos lados entre

2

2 e^{\frac{1}{u} l n (\frac{1}{u})} = 1

Dividir ambos lados entre

2

\frac{2 e ^{\frac{1}{u} l n (\frac{1}{u})}}{2} = \frac{1}{2}

Simplificar

e^{\frac{1}{u} l n (\frac{1}{u})} = \frac{1}{2}

e^{\frac{1}{u} l n (\frac{1}{u})} = \frac{1}{2}

Simplificar

e^{\frac{1}{u} l n (\frac{1}{u})} = \frac{2}{2}

Aplicar las leyes de los exponentes

e^{\frac{1}{u} l n (\frac{1}{u})} = \frac{2}{2}

Convertir a base

2 : e^{\frac{1}{u} l n (\frac{1}{u})} = 2^{\frac{- 1}{2}}

Simplificar

e^{\frac{1}{u} l n (\frac{1}{u})} = 2^{\frac{- 1}{2}}

e^{\frac{1}{u} l n (\frac{1}{u})} = 2^{\frac{- 1}{2}}

f (x) = g (x)

, entonces

ln (f (x)) = ln (g (x))

ln (e^{\frac{1}{u} l n (\frac{1}{u})}) = ln (2^{\frac{- 1}{2}})

Aplicar las propiedades de los logaritmos:

ln (e^{a}) = a

ln (e^{\frac{1}{u} l n (\frac{1}{u})}) = \frac{1}{u} ln (\frac{1}{u})

\frac{1}{u} ln (\frac{1}{u}) = ln (2^{\frac{- 1}{2}})

Aplicar las propiedades de los logaritmos:

ln (x^{a}) = a \cdot ln (x)

ln (2^{\frac{- 1}{2}}) = \frac{- 1}{2} ln (2)

\frac{1}{u} ln (\frac{1}{u}) = \frac{- 1}{2} ln (2)

\frac{1}{u} ln (\frac{1}{u}) = \frac{- 1}{2} ln (2)

Resolver

\frac{1}{u} ln (\frac{1}{u}) = \frac{- 1}{2} ln (2) : u = 4, u = 2

\frac{1}{u} ln (\frac{1}{u}) = \frac{- 1}{2} ln (2)

Multiplicar ambos lados por

u

\frac{1}{u} ln (\frac{1}{u}) = \frac{- 1}{2} ln (2)

Multiplicar ambos lados por

u

\frac{1}{u} ln (\frac{1}{u}) u = \frac{- 1}{2} ln (2) u

Simplificar

\frac{1}{u} ln (\frac{1}{u}) u = \frac{- 1}{2} ln (2) u

Simplificar

\frac{1}{u} ln (\frac{1}{u}) u : ln (\frac{1}{u})

\frac{1}{u} ln (\frac{1}{u}) u

Multiplicar fracciones:

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

= \frac{1 \cdot ln ( \frac{1}{u} ) u}{u}

Eliminar los terminos comunes:

u

= 1 \cdot ln (\frac{1}{u})

Multiplicar:

1 \cdot ln (\frac{1}{u}) = ln (\frac{1}{u})

= ln (\frac{1}{u})

Simplificar

\frac{- 1}{2} ln (2) u : - \frac{1}{2} u ln (2)

\frac{- 1}{2} ln (2) u

Aplicar las propiedades de las fracciones:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{2} u ln (2)

ln (\frac{1}{u}) = - \frac{1}{2} u ln (2)

ln (\frac{1}{u}) = - \frac{1}{2} u ln (2)

ln (\frac{1}{u}) = - \frac{1}{2} u ln (2)

f (x) = g (x),

entonces

a^{f (x)} = a^{g (x)}

e^{l n (\frac{1}{u})} = e^{- \frac{1}{2} u l n (2)}

Simplificar

e^{l n (\frac{1}{u})} : \frac{1}{u}

e^{l n (\frac{1}{u})}

Aplicar las propiedades de los logaritmos:

a^{l o g_{a} (b)} = b

= \frac{1}{u}

Simplificar

e^{- \frac{1}{2} u l n (2)} : 2^{- \frac{1}{2} u}

e^{- \frac{1}{2} u l n (2)}

Aplicar las leyes de los exponentes:

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

= (e^{l n (2)})^{- \frac{1}{2} u}

Aplicar las propiedades de los logaritmos:

a^{l o g_{a} (b)} = b

e^{l n (2)} = 2

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

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

Multiplicar ambos lados por

u

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

Simplificar

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

Resolver

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

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

Preparar

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

para la forma de Lambert:

1 = e^{- \frac{1}{2} l n (2) u} u

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

x e^{x} = a

es una ecuación en forma de Lambert

Aplicar las leyes de los exponentes

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

Convertir a base

e : 1 = e^{l n (2) (- \frac{1}{2} u)} u

Aplicar las leyes de los exponentes:

a = b^{l o g_{b} (a)}

2^{- \frac{1}{2} u} = (e^{l n (2)})^{- \frac{1}{2} u}

1 = (e^{l n (2)})^{- \frac{1}{2} u} u

Aplicar las leyes de los exponentes:

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

(e^{l n (2)})^{- \frac{1}{2} u} = e^{l n (2) (- \frac{1}{2} u)}

1 = e^{l n (2) (- \frac{1}{2} u)} u

1 = e^{l n (2) (- \frac{1}{2} u)} u

Simplificar

1 = e^{- \frac{1}{2} l n (2) u} u

1 = e^{- \frac{1}{2} l n (2) u} u

Re-escribir la ecuación con

- \frac{1}{2} u ln (2) = v

u = - \frac{2 v}{ln ( 2 )}

1 = e^{v} (- \frac{2 v}{ln ( 2 )})

Reescribir

1 = e^{v} (- \frac{2 v}{ln ( 2 )})

en forma de Lambert:

e^{v} v = - \frac{ln ( 2 )}{2}

1 = e^{v} (- \frac{2 v}{ln ( 2 )})

x e^{x} = a

es una ecuación en forma de Lambert

Intercambiar lados

e^{v} (- \frac{2 v}{ln ( 2 )}) = 1

Multiplicar ambos lados por

ln (2)

e^{v} (- \frac{2 v}{ln ( 2 )}) ln (2) = 1 \cdot ln (2)

Simplificar

- 2 e^{v} v = ln (2)

Dividir ambos lados entre

- 2

\frac{- 2 e ^{v} v}{- 2} = \frac{ln ( 2 )}{- 2}

Simplificar

e^{v} v = - \frac{ln ( 2 )}{2}

Resolver

e^{v} v = - \frac{ln ( 2 )}{2} : v = - 2 ln (2), v = - ln (2)

e^{v} v = - \frac{ln ( 2 )}{2}

Las soluciones para

x e^{x} = a

donde

- \frac{1}{e} \leq a < 0

son ramas principales y negativas de la función

W

de Lambert:

x = W_{0} (a), W_{- 1} (a)

v = W_{- 1} (- \frac{ln ( 2 )}{2}), v = W_{0} (- \frac{ln ( 2 )}{2})

Simplificar

v = - 2 ln (2), v = - ln (2)

v = - 2 ln (2), v = - ln (2)

Sustituir hacia atrás la

v = - \frac{1}{2} u ln (2),

resolver para

u

Resolver

- \frac{1}{2} u ln (2) = - 2 ln (2) : u = 4

- \frac{1}{2} u ln (2) = - 2 ln (2)

Multiplicar ambos lados por

- 2

- \frac{1}{2} u ln (2) = - 2 ln (2)

Multiplicar ambos lados por

- 2

(- \frac{1}{2} u ln (2)) (- 2) = (- 2 ln (2)) (- 2)

Simplificar

ln (2) u = 4 ln (2)

ln (2) u = 4 ln (2)

Dividir ambos lados entre

ln (2)

ln (2) u = 4 ln (2)

Dividir ambos lados entre

ln (2)

\frac{ln ( 2 ) u}{ln ( 2 )} = \frac{4 ln ( 2 )}{ln ( 2 )}

Simplificar

u = 4

u = 4

Resolver

- \frac{1}{2} u ln (2) = - ln (2) : u = 2

- \frac{1}{2} u ln (2) = - ln (2)

Multiplicar ambos lados por

- 2

- \frac{1}{2} u ln (2) = - ln (2)

Multiplicar ambos lados por

- 2

(- \frac{1}{2} u ln (2)) (- 2) = (- ln (2)) (- 2)

Simplificar

ln (2) u = 2 ln (2)

ln (2) u = 2 ln (2)

Dividir ambos lados entre

ln (2)

ln (2) u = 2 ln (2)

Dividir ambos lados entre

ln (2)

\frac{ln ( 2 ) u}{ln ( 2 )} = \frac{2 ln ( 2 )}{ln ( 2 )}

Simplificar

u = 2

u = 2

u = 4, u = 2

u = 4, u = 2

Verificar las soluciones:

u = 4

Verdadero

, u = 2

Verdadero

Verificar las soluciones sustituyéndolas en

\frac{1}{u} ln (\frac{1}{u}) = \frac{- 1}{2} ln (2)

Quitar las que no concuerden con la ecuación.

Sustituir

u = 4 :

Verdadero

\frac{1}{4} ln (\frac{1}{4}) = \frac{- 1}{2} ln (2)

\frac{1}{4} ln (\frac{1}{4}) = - \frac{1}{2} ln (2)

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

Simplificar

ln (\frac{1}{4}) : - 2 ln (2)

ln (\frac{1}{4})

Aplicar las propiedades de los logaritmos:

lo g_{a} (\frac{1}{x}) = - lo g_{a} (x)

= - ln (4)

Reescribir

4

utilizando potencias:

4 = 2^{2}

= - ln (2^{2})

Aplicar las propiedades de los logaritmos:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

ln (2^{2}) = 2 ln (2)

= - 2 ln (2)

= \frac{1}{4} (- 2 ln (2))

Quitar los parentesis:

(- a) = - a

= - \frac{1}{4} \cdot 2 ln (2)

Multiplicar fracciones:

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

= - \frac{1 \cdot 2}{4} ln (2)

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

\frac{1 \cdot 2}{4}

Multiplicar los numeros:

1 \cdot 2 = 2

= \frac{2}{4}

Eliminar los terminos comunes:

2

= \frac{1}{2}

= - \frac{1}{2} ln (2)

\frac{- 1}{2} ln (2) = - \frac{1}{2} ln (2)

\frac{- 1}{2} ln (2)

Aplicar las propiedades de las fracciones:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{2} ln (2)

- \frac{1}{2} ln (2) = - \frac{1}{2} ln (2)

Verdadero

Sustituir

u = 2 :

Verdadero

\frac{1}{2} ln (\frac{1}{2}) = \frac{- 1}{2} ln (2)

\frac{1}{2} ln (\frac{1}{2}) = - \frac{1}{2} ln (2)

\frac{1}{2} ln (\frac{1}{2})

Simplificar

ln (\frac{1}{2}) : - ln (2)

ln (\frac{1}{2})

Aplicar las propiedades de los logaritmos:

lo g_{a} (\frac{1}{x}) = - lo g_{a} (x)

= - ln (2)

= \frac{1}{2} (- ln (2))

Quitar los parentesis:

(- a) = - a

= - \frac{1}{2} ln (2)

\frac{- 1}{2} ln (2) = - \frac{1}{2} ln (2)

\frac{- 1}{2} ln (2)

Aplicar las propiedades de las fracciones:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{2} ln (2)

- \frac{1}{2} ln (2) = - \frac{1}{2} ln (2)

Verdadero

Las soluciones son

u = 4, u = 2

u = 4, u = 2

Sustituir en la ecuación

u = csc (x)

csc (x) = 4, csc (x) = 2

csc (x) = 4, csc (x) = 2

csc (x) = 4 : x = arccsc (4) + 2 πn, x = π - arccsc (4) + 2 πn

csc (x) = 4

Aplicar propiedades trigonométricas inversas

csc (x) = 4

Soluciones generales para

csc (x) = 4

csc (x) = a \Rightarrow x = arccsc (a) + 2 πn, x = π - arccsc (a) + 2 πn

x = arccsc (4) + 2 πn, x = π - arccsc (4) + 2 πn

x = arccsc (4) + 2 πn, x = π - arccsc (4) + 2 πn

csc (x) = 2 : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

csc (x) = 2

Soluciones generales para

csc (x) = 2

csc (x)

tabla de valores periódicos con

2 πn

intervalos:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Combinar toda las soluciones

x = arccsc (4) + 2 πn, x = π - arccsc (4) + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Mostrar soluciones en forma decimal

x = 0.25268 \dots + 2 πn, x = π - 0.25268 \dots + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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