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Popular Trigonometría >

csc^2(x)-3=6tan(x)

  • Pre-Álgebra
  • Álgebra
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Solución

csc2(x)−3=6tan(x)

Solución

x=0.43008…+πn
+1
Grados
x=24.64185…∘+180∘n
Pasos de solución
csc2(x)−3=6tan(x)
Elevar al cuadrado ambos lados(csc2(x)−3)2=(6tan(x))2
Restar (6tan(x))2 de ambos lados(csc2(x)−3)2−36tan2(x)=0
Re-escribir usando identidades trigonométricas
(−3+csc2(x))2−36tan2(x)
Utilizar la identidad pitagórica: csc2(x)=1+cot2(x)=(−3+1+cot2(x))2−36tan2(x)
Simplificar=(cot2(x)−2)2−36tan2(x)
(−2+cot2(x))2−36tan2(x)=0
Factorizar (−2+cot2(x))2−36tan2(x):(−2+cot2(x)+6tan(x))(−2+cot2(x)−6tan(x))
(−2+cot2(x))2−36tan2(x)
Reescribir (−2+cot2(x))2−36tan2(x) como (−2+cot2(x))2−(6tan(x))2
(−2+cot2(x))2−36tan2(x)
Reescribir 36 como 62=(−2+cot2(x))2−62tan2(x)
Aplicar las leyes de los exponentes: ambm=(ab)m62tan2(x)=(6tan(x))2=(−2+cot2(x))2−(6tan(x))2
=(−2+cot2(x))2−(6tan(x))2
Aplicar la siguiente regla para binomios al cuadrado: x2−y2=(x+y)(x−y)(−2+cot2(x))2−(6tan(x))2=((−2+cot2(x))+6tan(x))((−2+cot2(x))−6tan(x))=((−2+cot2(x))+6tan(x))((−2+cot2(x))−6tan(x))
Simplificar=(cot2(x)+6tan(x)−2)(cot2(x)−6tan(x)−2)
(−2+cot2(x)+6tan(x))(−2+cot2(x)−6tan(x))=0
Resolver cada parte por separado−2+cot2(x)+6tan(x)=0or−2+cot2(x)−6tan(x)=0
−2+cot2(x)+6tan(x)=0:x=arccot(−2.17998…)+πn
−2+cot2(x)+6tan(x)=0
Re-escribir usando identidades trigonométricas
−2+cot2(x)+6tan(x)
Utilizar la identidad trigonométrica básica: tan(x)=cot(x)1​=−2+cot2(x)+6⋅cot(x)1​
6⋅cot(x)1​=cot(x)6​
6⋅cot(x)1​
Multiplicar fracciones: a⋅cb​=ca⋅b​=cot(x)1⋅6​
Multiplicar los numeros: 1⋅6=6=cot(x)6​
=−2+cot2(x)+cot(x)6​
−2+cot2(x)+cot(x)6​=0
Usando el método de sustitución
−2+cot2(x)+cot(x)6​=0
Sea: cot(x)=u−2+u2+u6​=0
−2+u2+u6​=0:u≈−2.17998…
−2+u2+u6​=0
Multiplicar ambos lados por u
−2+u2+u6​=0
Multiplicar ambos lados por u−2u+u2u+u6​u=0⋅u
Simplificar
−2u+u2u+u6​u=0⋅u
Simplificar u2u:u3
u2u
Aplicar las leyes de los exponentes: ab⋅ac=ab+cu2u=u2+1=u2+1
Sumar: 2+1=3=u3
Simplificar u6​u:6
u6​u
Multiplicar fracciones: a⋅cb​=ca⋅b​=u6u​
Eliminar los terminos comunes: u=6
Simplificar 0⋅u:0
0⋅u
Aplicar la regla 0⋅a=0=0
−2u+u3+6=0
−2u+u3+6=0
−2u+u3+6=0
Resolver −2u+u3+6=0:u≈−2.17998…
−2u+u3+6=0
Escribir en la forma binómica an​xn+…+a1​x+a0​=0u3−2u+6=0
Encontrar una solución para u3−2u+6=0 utilizando el método de Newton-Raphson:u≈−2.17998…
u3−2u+6=0
Definición del método de Newton-Raphson
f(u)=u3−2u+6
Hallar f′(u):3u2−2
dud​(u3−2u+6)
Aplicar la regla de la suma/diferencia: (f±g)′=f′±g′=dud​(u3)−dud​(2u)+dud​(6)
dud​(u3)=3u2
dud​(u3)
Aplicar la regla de la potencia: dxd​(xa)=a⋅xa−1=3u3−1
Simplificar=3u2
dud​(2u)=2
dud​(2u)
Sacar la constante: (a⋅f)′=a⋅f′=2dudu​
Aplicar la regla de derivación: dudu​=1=2⋅1
Simplificar=2
dud​(6)=0
dud​(6)
Derivada de una constante: dxd​(a)=0=0
=3u2−2+0
Simplificar=3u2−2
Sea u0​=3Calcular un+1​ hasta que Δun+1​<0.000001
u1​=1.92:Δu1​=1.08
f(u0​)=33−2⋅3+6=27f′(u0​)=3⋅32−2=25u1​=1.92
Δu1​=∣1.92−3∣=1.08Δu1​=1.08
u2​=0.90027…:Δu2​=1.01972…
f(u1​)=1.923−2⋅1.92+6=9.237888f′(u1​)=3⋅1.922−2=9.0592u2​=0.90027…
Δu2​=∣0.90027…−1.92∣=1.01972…Δu2​=1.01972…
u3​=−10.52325…:Δu3​=11.42353…
f(u2​)=0.90027…3−2⋅0.90027…+6=4.92911…f′(u2​)=3⋅0.90027…2−2=0.43148…u3​=−10.52325…
Δu3​=∣−10.52325…−0.90027…∣=11.42353…Δu3​=11.42353…
u4​=−7.07616…:Δu4​=3.44709…
f(u3​)=(−10.52325…)3−2(−10.52325…)+6=−1138.28856…f′(u3​)=3(−10.52325…)2−2=330.21696…u4​=−7.07616…
Δu4​=∣−7.07616…−(−10.52325…)∣=3.44709…Δu4​=3.44709…
u5​=−4.82158…:Δu5​=2.25458…
f(u4​)=(−7.07616…)3−2(−7.07616…)+6=−334.16639…f′(u4​)=3(−7.07616…)2−2=148.21639…u5​=−4.82158…
Δu5​=∣−4.82158…−(−7.07616…)∣=2.25458…Δu5​=2.25458…
u6​=−3.39785…:Δu6​=1.42372…
f(u5​)=(−4.82158…)3−2(−4.82158…)+6=−96.44728…f′(u5​)=3(−4.82158…)2−2=67.74295…u6​=−3.39785…
Δu6​=∣−3.39785…−(−4.82158…)∣=1.42372…Δu6​=1.42372…
u7​=−2.58789…:Δu7​=0.80995…
f(u6​)=(−3.39785…)3−2(−3.39785…)+6=−26.43402…f′(u6​)=3(−3.39785…)2−2=32.63630…u7​=−2.58789…
Δu7​=∣−2.58789…−(−3.39785…)∣=0.80995…Δu7​=0.80995…
u8​=−2.24763…:Δu8​=0.34026…
f(u7​)=(−2.58789…)3−2(−2.58789…)+6=−6.15594…f′(u7​)=3(−2.58789…)2−2=18.09167…u8​=−2.24763…
Δu8​=∣−2.24763…−(−2.58789…)∣=0.34026…Δu8​=0.34026…
u9​=−2.18230…:Δu9​=0.06533…
f(u8​)=(−2.24763…)3−2(−2.24763…)+6=−0.85948…f′(u8​)=3(−2.24763…)2−2=13.15559…u9​=−2.18230…
Δu9​=∣−2.18230…−(−2.24763…)∣=0.06533…Δu9​=0.06533…
u10​=−2.17998…:Δu10​=0.00231…
f(u9​)=(−2.18230…)3−2(−2.18230…)+6=−0.02850…f′(u9​)=3(−2.18230…)2−2=12.28734…u10​=−2.17998…
Δu10​=∣−2.17998…−(−2.18230…)∣=0.00231…Δu10​=0.00231…
u11​=−2.17998…:Δu11​=2.87294E−6
f(u10​)=(−2.17998…)3−2(−2.17998…)+6=−0.00003…f′(u10​)=3(−2.17998…)2−2=12.25699…u11​=−2.17998…
Δu11​=∣−2.17998…−(−2.17998…)∣=2.87294E−6Δu11​=2.87294E−6
u12​=−2.17998…:Δu12​=4.4041E−12
f(u11​)=(−2.17998…)3−2(−2.17998…)+6=−5.39808E−11f′(u11​)=3(−2.17998…)2−2=12.25695…u12​=−2.17998…
Δu12​=∣−2.17998…−(−2.17998…)∣=4.4041E−12Δu12​=4.4041E−12
u≈−2.17998…
Aplicar la división larga Equation0:u+2.17998…u3−2u+6​=u2−2.17998…u+2.75231…
u2−2.17998…u+2.75231…≈0
Encontrar una solución para u2−2.17998…u+2.75231…=0 utilizando el método de Newton-Raphson:Sin solución para u∈R
u2−2.17998…u+2.75231…=0
Definición del método de Newton-Raphson
f(u)=u2−2.17998…u+2.75231…
Hallar f′(u):2u−2.17998…
dud​(u2−2.17998…u+2.75231…)
Aplicar la regla de la suma/diferencia: (f±g)′=f′±g′=dud​(u2)−dud​(2.17998…u)+dud​(2.75231…)
dud​(u2)=2u
dud​(u2)
Aplicar la regla de la potencia: dxd​(xa)=a⋅xa−1=2u2−1
Simplificar=2u
dud​(2.17998…u)=2.17998…
dud​(2.17998…u)
Sacar la constante: (a⋅f)′=a⋅f′=2.17998…dudu​
Aplicar la regla de derivación: dudu​=1=2.17998…⋅1
Simplificar=2.17998…
dud​(2.75231…)=0
dud​(2.75231…)
Derivada de una constante: dxd​(a)=0=0
=2u−2.17998…+0
Simplificar=2u−2.17998…
Sea u0​=1Calcular un+1​ hasta que Δun+1​<0.000001
u1​=9.73612…:Δu1​=8.73612…
f(u0​)=12−2.17998…⋅1+2.75231…=1.57233…f′(u0​)=2⋅1−2.17998…=−0.17998…u1​=9.73612…
Δu1​=∣9.73612…−1∣=8.73612…Δu1​=8.73612…
u2​=5.32259…:Δu2​=4.41352…
f(u1​)=9.73612…2−2.17998…⋅9.73612…+2.75231…=76.31980…f′(u1​)=2⋅9.73612…−2.17998…=17.29226…u2​=5.32259…
Δu2​=∣5.32259…−9.73612…∣=4.41352…Δu2​=4.41352…
u3​=3.02150…:Δu3​=2.30108…
f(u2​)=5.32259…2−2.17998…⋅5.32259…+2.75231…=19.47919…f′(u2​)=2⋅5.32259…−2.17998…=8.46521…u3​=3.02150…
Δu3​=∣3.02150…−5.32259…∣=2.30108…Δu3​=2.30108…
u4​=1.65082…:Δu4​=1.37068…
f(u3​)=3.02150…2−2.17998…⋅3.02150…+2.75231…=5.29500…f′(u3​)=2⋅3.02150…−2.17998…=3.86303…u4​=1.65082…
Δu4​=∣1.65082…−3.02150…∣=1.37068…Δu4​=1.37068…
u5​=−0.02415…:Δu5​=1.67497…
f(u4​)=1.65082…2−2.17998…⋅1.65082…+2.75231…=1.87877…f′(u4​)=2⋅1.65082…−2.17998…=1.12167…u5​=−0.02415…
Δu5​=∣−0.02415…−1.65082…∣=1.67497…Δu5​=1.67497…
u6​=1.23490…:Δu6​=1.25906…
f(u5​)=(−0.02415…)2−2.17998…(−0.02415…)+2.75231…=2.80555…f′(u5​)=2(−0.02415…)−2.17998…=−2.22828…u6​=1.23490…
Δu6​=∣1.23490…−(−0.02415…)∣=1.25906…Δu6​=1.25906…
u7​=−4.23449…:Δu7​=5.46939…
f(u6​)=1.23490…2−2.17998…⋅1.23490…+2.75231…=1.58523…f′(u6​)=2⋅1.23490…−2.17998…=0.28983…u7​=−4.23449…
Δu7​=∣−4.23449…−1.23490…∣=5.46939…Δu7​=5.46939…
u8​=−1.42535…:Δu8​=2.80913…
f(u7​)=(−4.23449…)2−2.17998…(−4.23449…)+2.75231…=29.91433…f′(u7​)=2(−4.23449…)−2.17998…=−10.64896…u8​=−1.42535…
Δu8​=∣−1.42535…−(−4.23449…)∣=2.80913…Δu8​=2.80913…
u9​=0.14325…:Δu9​=1.56861…
f(u8​)=(−1.42535…)2−2.17998…(−1.42535…)+2.75231…=7.89122…f′(u8​)=2(−1.42535…)−2.17998…=−5.03069…u9​=0.14325…
Δu9​=∣0.14325…−(−1.42535…)∣=1.56861…Δu9​=1.56861…
u10​=1.44274…:Δu10​=1.29948…
f(u9​)=0.14325…2−2.17998…⋅0.14325…+2.75231…=2.46054…f′(u9​)=2⋅0.14325…−2.17998…=−1.89347…u10​=1.44274…
Δu10​=∣1.44274…−0.14325…∣=1.29948…Δu10​=1.29948…
u11​=−0.95081…:Δu11​=2.39356…
f(u10​)=1.44274…2−2.17998…⋅1.44274…+2.75231…=1.68867…f′(u10​)=2⋅1.44274…−2.17998…=0.70550…u11​=−0.95081…
Δu11​=∣−0.95081…−1.44274…∣=2.39356…Δu11​=2.39356…
u12​=0.45282…:Δu12​=1.40364…
f(u11​)=(−0.95081…)2−2.17998…(−0.95081…)+2.75231…=5.72913…f′(u11​)=2(−0.95081…)−2.17998…=−4.08161…u12​=0.45282…
Δu12​=∣0.45282…−(−0.95081…)∣=1.40364…Δu12​=1.40364…
u13​=1.99891…:Δu13​=1.54608…
f(u12​)=0.45282…2−2.17998…⋅0.45282…+2.75231…=1.97021…f′(u12​)=2⋅0.45282…−2.17998…=−1.27432…u13​=1.99891…
Δu13​=∣1.99891…−0.45282…∣=1.54608…Δu13​=1.54608…
No se puede encontrar solución
La solución esu≈−2.17998…
u≈−2.17998…
Verificar las soluciones
Encontrar los puntos no definidos (singularidades):u=0
Tomar el(los) denominador(es) de −2+u2+u6​ y comparar con cero
u=0
Los siguientes puntos no están definidosu=0
Combinar los puntos no definidos con las soluciones:
u≈−2.17998…
Sustituir en la ecuación u=cot(x)cot(x)≈−2.17998…
cot(x)≈−2.17998…
cot(x)=−2.17998…:x=arccot(−2.17998…)+πn
cot(x)=−2.17998…
Aplicar propiedades trigonométricas inversas
cot(x)=−2.17998…
Soluciones generales para cot(x)=−2.17998…cot(x)=−a⇒x=arccot(−a)+πnx=arccot(−2.17998…)+πn
x=arccot(−2.17998…)+πn
Combinar toda las solucionesx=arccot(−2.17998…)+πn
−2+cot2(x)−6tan(x)=0:x=arccot(2.17998…)+πn
−2+cot2(x)−6tan(x)=0
Re-escribir usando identidades trigonométricas
−2+cot2(x)−6tan(x)
Utilizar la identidad trigonométrica básica: tan(x)=cot(x)1​=−2+cot2(x)−6⋅cot(x)1​
6⋅cot(x)1​=cot(x)6​
6⋅cot(x)1​
Multiplicar fracciones: a⋅cb​=ca⋅b​=cot(x)1⋅6​
Multiplicar los numeros: 1⋅6=6=cot(x)6​
=−2+cot2(x)−cot(x)6​
−2+cot2(x)−cot(x)6​=0
Usando el método de sustitución
−2+cot2(x)−cot(x)6​=0
Sea: cot(x)=u−2+u2−u6​=0
−2+u2−u6​=0:u≈2.17998…
−2+u2−u6​=0
Multiplicar ambos lados por u
−2+u2−u6​=0
Multiplicar ambos lados por u−2u+u2u−u6​u=0⋅u
Simplificar
−2u+u2u−u6​u=0⋅u
Simplificar u2u:u3
u2u
Aplicar las leyes de los exponentes: ab⋅ac=ab+cu2u=u2+1=u2+1
Sumar: 2+1=3=u3
Simplificar −u6​u:−6
−u6​u
Multiplicar fracciones: a⋅cb​=ca⋅b​=−u6u​
Eliminar los terminos comunes: u=−6
Simplificar 0⋅u:0
0⋅u
Aplicar la regla 0⋅a=0=0
−2u+u3−6=0
−2u+u3−6=0
−2u+u3−6=0
Resolver −2u+u3−6=0:u≈2.17998…
−2u+u3−6=0
Escribir en la forma binómica an​xn+…+a1​x+a0​=0u3−2u−6=0
Encontrar una solución para u3−2u−6=0 utilizando el método de Newton-Raphson:u≈2.17998…
u3−2u−6=0
Definición del método de Newton-Raphson
f(u)=u3−2u−6
Hallar f′(u):3u2−2
dud​(u3−2u−6)
Aplicar la regla de la suma/diferencia: (f±g)′=f′±g′=dud​(u3)−dud​(2u)−dud​(6)
dud​(u3)=3u2
dud​(u3)
Aplicar la regla de la potencia: dxd​(xa)=a⋅xa−1=3u3−1
Simplificar=3u2
dud​(2u)=2
dud​(2u)
Sacar la constante: (a⋅f)′=a⋅f′=2dudu​
Aplicar la regla de derivación: dudu​=1=2⋅1
Simplificar=2
dud​(6)=0
dud​(6)
Derivada de una constante: dxd​(a)=0=0
=3u2−2−0
Simplificar=3u2−2
Sea u0​=−3Calcular un+1​ hasta que Δun+1​<0.000001
u1​=−1.92:Δu1​=1.08
f(u0​)=(−3)3−2(−3)−6=−27f′(u0​)=3(−3)2−2=25u1​=−1.92
Δu1​=∣−1.92−(−3)∣=1.08Δu1​=1.08
u2​=−0.90027…:Δu2​=1.01972…
f(u1​)=(−1.92)3−2(−1.92)−6=−9.237888f′(u1​)=3(−1.92)2−2=9.0592u2​=−0.90027…
Δu2​=∣−0.90027…−(−1.92)∣=1.01972…Δu2​=1.01972…
u3​=10.52325…:Δu3​=11.42353…
f(u2​)=(−0.90027…)3−2(−0.90027…)−6=−4.92911…f′(u2​)=3(−0.90027…)2−2=0.43148…u3​=10.52325…
Δu3​=∣10.52325…−(−0.90027…)∣=11.42353…Δu3​=11.42353…
u4​=7.07616…:Δu4​=3.44709…
f(u3​)=10.52325…3−2⋅10.52325…−6=1138.28856…f′(u3​)=3⋅10.52325…2−2=330.21696…u4​=7.07616…
Δu4​=∣7.07616…−10.52325…∣=3.44709…Δu4​=3.44709…
u5​=4.82158…:Δu5​=2.25458…
f(u4​)=7.07616…3−2⋅7.07616…−6=334.16639…f′(u4​)=3⋅7.07616…2−2=148.21639…u5​=4.82158…
Δu5​=∣4.82158…−7.07616…∣=2.25458…Δu5​=2.25458…
u6​=3.39785…:Δu6​=1.42372…
f(u5​)=4.82158…3−2⋅4.82158…−6=96.44728…f′(u5​)=3⋅4.82158…2−2=67.74295…u6​=3.39785…
Δu6​=∣3.39785…−4.82158…∣=1.42372…Δu6​=1.42372…
u7​=2.58789…:Δu7​=0.80995…
f(u6​)=3.39785…3−2⋅3.39785…−6=26.43402…f′(u6​)=3⋅3.39785…2−2=32.63630…u7​=2.58789…
Δu7​=∣2.58789…−3.39785…∣=0.80995…Δu7​=0.80995…
u8​=2.24763…:Δu8​=0.34026…
f(u7​)=2.58789…3−2⋅2.58789…−6=6.15594…f′(u7​)=3⋅2.58789…2−2=18.09167…u8​=2.24763…
Δu8​=∣2.24763…−2.58789…∣=0.34026…Δu8​=0.34026…
u9​=2.18230…:Δu9​=0.06533…
f(u8​)=2.24763…3−2⋅2.24763…−6=0.85948…f′(u8​)=3⋅2.24763…2−2=13.15559…u9​=2.18230…
Δu9​=∣2.18230…−2.24763…∣=0.06533…Δu9​=0.06533…
u10​=2.17998…:Δu10​=0.00231…
f(u9​)=2.18230…3−2⋅2.18230…−6=0.02850…f′(u9​)=3⋅2.18230…2−2=12.28734…u10​=2.17998…
Δu10​=∣2.17998…−2.18230…∣=0.00231…Δu10​=0.00231…
u11​=2.17998…:Δu11​=2.87294E−6
f(u10​)=2.17998…3−2⋅2.17998…−6=0.00003…f′(u10​)=3⋅2.17998…2−2=12.25699…u11​=2.17998…
Δu11​=∣2.17998…−2.17998…∣=2.87294E−6Δu11​=2.87294E−6
u12​=2.17998…:Δu12​=4.4041E−12
f(u11​)=2.17998…3−2⋅2.17998…−6=5.39808E−11f′(u11​)=3⋅2.17998…2−2=12.25695…u12​=2.17998…
Δu12​=∣2.17998…−2.17998…∣=4.4041E−12Δu12​=4.4041E−12
u≈2.17998…
Aplicar la división larga Equation0:u−2.17998…u3−2u−6​=u2+2.17998…u+2.75231…
u2+2.17998…u+2.75231…≈0
Encontrar una solución para u2+2.17998…u+2.75231…=0 utilizando el método de Newton-Raphson:Sin solución para u∈R
u2+2.17998…u+2.75231…=0
Definición del método de Newton-Raphson
f(u)=u2+2.17998…u+2.75231…
Hallar f′(u):2u+2.17998…
dud​(u2+2.17998…u+2.75231…)
Aplicar la regla de la suma/diferencia: (f±g)′=f′±g′=dud​(u2)+dud​(2.17998…u)+dud​(2.75231…)
dud​(u2)=2u
dud​(u2)
Aplicar la regla de la potencia: dxd​(xa)=a⋅xa−1=2u2−1
Simplificar=2u
dud​(2.17998…u)=2.17998…
dud​(2.17998…u)
Sacar la constante: (a⋅f)′=a⋅f′=2.17998…dudu​
Aplicar la regla de derivación: dudu​=1=2.17998…⋅1
Simplificar=2.17998…
dud​(2.75231…)=0
dud​(2.75231…)
Derivada de una constante: dxd​(a)=0=0
=2u+2.17998…+0
Simplificar=2u+2.17998…
Sea u0​=−1Calcular un+1​ hasta que Δun+1​<0.000001
u1​=−9.73612…:Δu1​=8.73612…
f(u0​)=(−1)2+2.17998…(−1)+2.75231…=1.57233…f′(u0​)=2(−1)+2.17998…=0.17998…u1​=−9.73612…
Δu1​=∣−9.73612…−(−1)∣=8.73612…Δu1​=8.73612…
u2​=−5.32259…:Δu2​=4.41352…
f(u1​)=(−9.73612…)2+2.17998…(−9.73612…)+2.75231…=76.31980…f′(u1​)=2(−9.73612…)+2.17998…=−17.29226…u2​=−5.32259…
Δu2​=∣−5.32259…−(−9.73612…)∣=4.41352…Δu2​=4.41352…
u3​=−3.02150…:Δu3​=2.30108…
f(u2​)=(−5.32259…)2+2.17998…(−5.32259…)+2.75231…=19.47919…f′(u2​)=2(−5.32259…)+2.17998…=−8.46521…u3​=−3.02150…
Δu3​=∣−3.02150…−(−5.32259…)∣=2.30108…Δu3​=2.30108…
u4​=−1.65082…:Δu4​=1.37068…
f(u3​)=(−3.02150…)2+2.17998…(−3.02150…)+2.75231…=5.29500…f′(u3​)=2(−3.02150…)+2.17998…=−3.86303…u4​=−1.65082…
Δu4​=∣−1.65082…−(−3.02150…)∣=1.37068…Δu4​=1.37068…
u5​=0.02415…:Δu5​=1.67497…
f(u4​)=(−1.65082…)2+2.17998…(−1.65082…)+2.75231…=1.87877…f′(u4​)=2(−1.65082…)+2.17998…=−1.12167…u5​=0.02415…
Δu5​=∣0.02415…−(−1.65082…)∣=1.67497…Δu5​=1.67497…
u6​=−1.23490…:Δu6​=1.25906…
f(u5​)=0.02415…2+2.17998…⋅0.02415…+2.75231…=2.80555…f′(u5​)=2⋅0.02415…+2.17998…=2.22828…u6​=−1.23490…
Δu6​=∣−1.23490…−0.02415…∣=1.25906…Δu6​=1.25906…
u7​=4.23449…:Δu7​=5.46939…
f(u6​)=(−1.23490…)2+2.17998…(−1.23490…)+2.75231…=1.58523…f′(u6​)=2(−1.23490…)+2.17998…=−0.28983…u7​=4.23449…
Δu7​=∣4.23449…−(−1.23490…)∣=5.46939…Δu7​=5.46939…
u8​=1.42535…:Δu8​=2.80913…
f(u7​)=4.23449…2+2.17998…⋅4.23449…+2.75231…=29.91433…f′(u7​)=2⋅4.23449…+2.17998…=10.64896…u8​=1.42535…
Δu8​=∣1.42535…−4.23449…∣=2.80913…Δu8​=2.80913…
u9​=−0.14325…:Δu9​=1.56861…
f(u8​)=1.42535…2+2.17998…⋅1.42535…+2.75231…=7.89122…f′(u8​)=2⋅1.42535…+2.17998…=5.03069…u9​=−0.14325…
Δu9​=∣−0.14325…−1.42535…∣=1.56861…Δu9​=1.56861…
u10​=−1.44274…:Δu10​=1.29948…
f(u9​)=(−0.14325…)2+2.17998…(−0.14325…)+2.75231…=2.46054…f′(u9​)=2(−0.14325…)+2.17998…=1.89347…u10​=−1.44274…
Δu10​=∣−1.44274…−(−0.14325…)∣=1.29948…Δu10​=1.29948…
u11​=0.95081…:Δu11​=2.39356…
f(u10​)=(−1.44274…)2+2.17998…(−1.44274…)+2.75231…=1.68867…f′(u10​)=2(−1.44274…)+2.17998…=−0.70550…u11​=0.95081…
Δu11​=∣0.95081…−(−1.44274…)∣=2.39356…Δu11​=2.39356…
u12​=−0.45282…:Δu12​=1.40364…
f(u11​)=0.95081…2+2.17998…⋅0.95081…+2.75231…=5.72913…f′(u11​)=2⋅0.95081…+2.17998…=4.08161…u12​=−0.45282…
Δu12​=∣−0.45282…−0.95081…∣=1.40364…Δu12​=1.40364…
u13​=−1.99891…:Δu13​=1.54608…
f(u12​)=(−0.45282…)2+2.17998…(−0.45282…)+2.75231…=1.97021…f′(u12​)=2(−0.45282…)+2.17998…=1.27432…u13​=−1.99891…
Δu13​=∣−1.99891…−(−0.45282…)∣=1.54608…Δu13​=1.54608…
No se puede encontrar solución
La solución esu≈2.17998…
u≈2.17998…
Verificar las soluciones
Encontrar los puntos no definidos (singularidades):u=0
Tomar el(los) denominador(es) de −2+u2−u6​ y comparar con cero
u=0
Los siguientes puntos no están definidosu=0
Combinar los puntos no definidos con las soluciones:
u≈2.17998…
Sustituir en la ecuación u=cot(x)cot(x)≈2.17998…
cot(x)≈2.17998…
cot(x)=2.17998…:x=arccot(2.17998…)+πn
cot(x)=2.17998…
Aplicar propiedades trigonométricas inversas
cot(x)=2.17998…
Soluciones generales para cot(x)=2.17998…cot(x)=a⇒x=arccot(a)+πnx=arccot(2.17998…)+πn
x=arccot(2.17998…)+πn
Combinar toda las solucionesx=arccot(2.17998…)+πn
Combinar toda las solucionesx=arccot(−2.17998…)+πn,x=arccot(2.17998…)+πn
Verificar las soluciones sustituyendo en la ecuación original
Verificar las soluciones sustituyéndolas en csc2(x)−3=6tan(x)
Quitar las que no concuerden con la ecuación.
Verificar la solución arccot(−2.17998…)+πn:Falso
arccot(−2.17998…)+πn
Sustituir n=1arccot(−2.17998…)+π1
Multiplicar csc2(x)−3=6tan(x) por x=arccot(−2.17998…)+π1csc2(arccot(−2.17998…)+π1)−3=6tan(arccot(−2.17998…)+π1)
Simplificar2.75231…=−2.75231…
⇒Falso
Verificar la solución arccot(2.17998…)+πn:Verdadero
arccot(2.17998…)+πn
Sustituir n=1arccot(2.17998…)+π1
Multiplicar csc2(x)−3=6tan(x) por x=arccot(2.17998…)+π1csc2(arccot(2.17998…)+π1)−3=6tan(arccot(2.17998…)+π1)
Simplificar2.75231…=2.75231…
⇒Verdadero
x=arccot(2.17998…)+πn
Mostrar soluciones en forma decimalx=0.43008…+πn

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