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Popular Trigonometría >

12=3sec(θ)+5csc(θ)

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Solución

12=3sec(θ)+5csc(θ)

Solución

θ=π−0.33570…+2πn,θ=1.07609…+2πn,θ=0.65383…+2πn,θ=−1.39422…+2πn
+1
Grados
θ=160.76534…∘+360∘n,θ=61.65590…∘+360∘n,θ=37.46199…∘+360∘n,θ=−79.88324…∘+360∘n
Pasos de solución
12=3sec(θ)+5csc(θ)
Restar 5csc(θ) de ambos lados3sec(θ)=12−5csc(θ)
Elevar al cuadrado ambos lados(3sec(θ))2=(12−5csc(θ))2
Restar (12−5csc(θ))2 de ambos lados9sec2(θ)−144+120csc(θ)−25csc2(θ)=0
Expresar con seno, coseno
−144+120csc(θ)−25csc2(θ)+9sec2(θ)
Utilizar la identidad trigonométrica básica: csc(x)=sin(x)1​=−144+120⋅sin(θ)1​−25(sin(θ)1​)2+9sec2(θ)
Utilizar la identidad trigonométrica básica: sec(x)=cos(x)1​=−144+120⋅sin(θ)1​−25(sin(θ)1​)2+9(cos(θ)1​)2
Simplificar −144+120⋅sin(θ)1​−25(sin(θ)1​)2+9(cos(θ)1​)2:cos2(θ)sin2(θ)−144cos2(θ)sin2(θ)+120cos2(θ)sin(θ)−25cos2(θ)+9sin2(θ)​
−144+120⋅sin(θ)1​−25(sin(θ)1​)2+9(cos(θ)1​)2
120⋅sin(θ)1​=sin(θ)120​
120⋅sin(θ)1​
Multiplicar fracciones: a⋅cb​=ca⋅b​=sin(θ)1⋅120​
Multiplicar los numeros: 1⋅120=120=sin(θ)120​
25(sin(θ)1​)2=sin2(θ)25​
25(sin(θ)1​)2
(sin(θ)1​)2=sin2(θ)1​
(sin(θ)1​)2
Aplicar las leyes de los exponentes: (ba​)c=bcac​=sin2(θ)12​
Aplicar la regla 1a=112=1=sin2(θ)1​
=25⋅sin2(θ)1​
Multiplicar fracciones: a⋅cb​=ca⋅b​=sin2(θ)1⋅25​
Multiplicar los numeros: 1⋅25=25=sin2(θ)25​
9(cos(θ)1​)2=cos2(θ)9​
9(cos(θ)1​)2
(cos(θ)1​)2=cos2(θ)1​
(cos(θ)1​)2
Aplicar las leyes de los exponentes: (ba​)c=bcac​=cos2(θ)12​
Aplicar la regla 1a=112=1=cos2(θ)1​
=9⋅cos2(θ)1​
Multiplicar fracciones: a⋅cb​=ca⋅b​=cos2(θ)1⋅9​
Multiplicar los numeros: 1⋅9=9=cos2(θ)9​
=−144+sin(θ)120​−sin2(θ)25​+cos2(θ)9​
Convertir a fracción: 144=cos2(θ)144cos2(θ)​=−cos2(θ)144cos2(θ)​+sin(θ)120​−sin2(θ)25​+cos2(θ)9​
Mínimo común múltiplo de cos2(θ),sin(θ),sin2(θ),cos2(θ):cos2(θ)sin2(θ)
cos2(θ),sin(θ),sin2(θ),cos2(θ)
Mínimo común múltiplo (MCM)
Calcular una expresión que este compuesta de factores que aparezcan en al menos una de las expresiones factorizadas=cos2(θ)sin2(θ)
Reescribir las fracciones basandose en el mínimo común denominador
Multiplicar cada numerador por la misma cantidad necesaria para multiplicar el denominador correspondiente y convertirlo en el mínimo común denominador
Para cos2(θ)144cos2(θ)​:multiplicar el denominador y el numerador por sin2(θ)cos2(θ)144cos2(θ)​=cos2(θ)sin2(θ)144cos2(θ)sin2(θ)​
Para sin(θ)120​:multiplicar el denominador y el numerador por cos2(θ)sin(θ)sin(θ)120​=sin(θ)cos2(θ)sin(θ)120cos2(θ)sin(θ)​=cos2(θ)sin2(θ)120cos2(θ)sin(θ)​
Para sin2(θ)25​:multiplicar el denominador y el numerador por cos2(θ)sin2(θ)25​=sin2(θ)cos2(θ)25cos2(θ)​
Para cos2(θ)9​:multiplicar el denominador y el numerador por sin2(θ)cos2(θ)9​=cos2(θ)sin2(θ)9sin2(θ)​
=−cos2(θ)sin2(θ)144cos2(θ)sin2(θ)​+cos2(θ)sin2(θ)120cos2(θ)sin(θ)​−sin2(θ)cos2(θ)25cos2(θ)​+cos2(θ)sin2(θ)9sin2(θ)​
Ya que los denominadores son iguales, combinar las fracciones: ca​±cb​=ca±b​=cos2(θ)sin2(θ)−144cos2(θ)sin2(θ)+120cos2(θ)sin(θ)−25cos2(θ)+9sin2(θ)​
=cos2(θ)sin2(θ)−144cos2(θ)sin2(θ)+120cos2(θ)sin(θ)−25cos2(θ)+9sin2(θ)​
cos2(θ)sin2(θ)−25cos2(θ)+9sin2(θ)+120cos2(θ)sin(θ)−144cos2(θ)sin2(θ)​=0
g(x)f(x)​=0⇒f(x)=0−25cos2(θ)+9sin2(θ)+120cos2(θ)sin(θ)−144cos2(θ)sin2(θ)=0
Re-escribir usando identidades trigonométricas
−25cos2(θ)+9sin2(θ)+120cos2(θ)sin(θ)−144cos2(θ)sin2(θ)
Utilizar la identidad pitagórica: cos2(x)+sin2(x)=1cos2(x)=1−sin2(x)=−25(1−sin2(θ))+9sin2(θ)+120(1−sin2(θ))sin(θ)−144(1−sin2(θ))sin2(θ)
Simplificar −25(1−sin2(θ))+9sin2(θ)+120(1−sin2(θ))sin(θ)−144(1−sin2(θ))sin2(θ):−110sin2(θ)+120sin(θ)−120sin3(θ)+144sin4(θ)−25
−25(1−sin2(θ))+9sin2(θ)+120(1−sin2(θ))sin(θ)−144(1−sin2(θ))sin2(θ)
=−25(1−sin2(θ))+9sin2(θ)+120sin(θ)(1−sin2(θ))−144sin2(θ)(1−sin2(θ))
Expandir −25(1−sin2(θ)):−25+25sin2(θ)
−25(1−sin2(θ))
Poner los parentesis utilizando: a(b−c)=ab−aca=−25,b=1,c=sin2(θ)=−25⋅1−(−25)sin2(θ)
Aplicar las reglas de los signos−(−a)=a=−25⋅1+25sin2(θ)
Multiplicar los numeros: 25⋅1=25=−25+25sin2(θ)
=−25+25sin2(θ)+9sin2(θ)+120(1−sin2(θ))sin(θ)−144(1−sin2(θ))sin2(θ)
Expandir 120sin(θ)(1−sin2(θ)):120sin(θ)−120sin3(θ)
120sin(θ)(1−sin2(θ))
Poner los parentesis utilizando: a(b−c)=ab−aca=120sin(θ),b=1,c=sin2(θ)=120sin(θ)⋅1−120sin(θ)sin2(θ)
=120⋅1⋅sin(θ)−120sin2(θ)sin(θ)
Simplificar 120⋅1⋅sin(θ)−120sin2(θ)sin(θ):120sin(θ)−120sin3(θ)
120⋅1⋅sin(θ)−120sin2(θ)sin(θ)
120⋅1⋅sin(θ)=120sin(θ)
120⋅1⋅sin(θ)
Multiplicar los numeros: 120⋅1=120=120sin(θ)
120sin2(θ)sin(θ)=120sin3(θ)
120sin2(θ)sin(θ)
Aplicar las leyes de los exponentes: ab⋅ac=ab+csin2(θ)sin(θ)=sin2+1(θ)=120sin2+1(θ)
Sumar: 2+1=3=120sin3(θ)
=120sin(θ)−120sin3(θ)
=120sin(θ)−120sin3(θ)
=−25+25sin2(θ)+9sin2(θ)+120sin(θ)−120sin3(θ)−144(1−sin2(θ))sin2(θ)
Expandir −144sin2(θ)(1−sin2(θ)):−144sin2(θ)+144sin4(θ)
−144sin2(θ)(1−sin2(θ))
Poner los parentesis utilizando: a(b−c)=ab−aca=−144sin2(θ),b=1,c=sin2(θ)=−144sin2(θ)⋅1−(−144sin2(θ))sin2(θ)
Aplicar las reglas de los signos−(−a)=a=−144⋅1⋅sin2(θ)+144sin2(θ)sin2(θ)
Simplificar −144⋅1⋅sin2(θ)+144sin2(θ)sin2(θ):−144sin2(θ)+144sin4(θ)
−144⋅1⋅sin2(θ)+144sin2(θ)sin2(θ)
144⋅1⋅sin2(θ)=144sin2(θ)
144⋅1⋅sin2(θ)
Multiplicar los numeros: 144⋅1=144=144sin2(θ)
144sin2(θ)sin2(θ)=144sin4(θ)
144sin2(θ)sin2(θ)
Aplicar las leyes de los exponentes: ab⋅ac=ab+csin2(θ)sin2(θ)=sin2+2(θ)=144sin2+2(θ)
Sumar: 2+2=4=144sin4(θ)
=−144sin2(θ)+144sin4(θ)
=−144sin2(θ)+144sin4(θ)
=−25+25sin2(θ)+9sin2(θ)+120sin(θ)−120sin3(θ)−144sin2(θ)+144sin4(θ)
Simplificar −25+25sin2(θ)+9sin2(θ)+120sin(θ)−120sin3(θ)−144sin2(θ)+144sin4(θ):−110sin2(θ)+120sin(θ)−120sin3(θ)+144sin4(θ)−25
−25+25sin2(θ)+9sin2(θ)+120sin(θ)−120sin3(θ)−144sin2(θ)+144sin4(θ)
Agrupar términos semejantes=25sin2(θ)+9sin2(θ)+120sin(θ)−120sin3(θ)−144sin2(θ)+144sin4(θ)−25
Sumar elementos similares: 25sin2(θ)+9sin2(θ)−144sin2(θ)=−110sin2(θ)=−110sin2(θ)+120sin(θ)−120sin3(θ)+144sin4(θ)−25
=−110sin2(θ)+120sin(θ)−120sin3(θ)+144sin4(θ)−25
=−110sin2(θ)+120sin(θ)−120sin3(θ)+144sin4(θ)−25
−25−110sin2(θ)+120sin(θ)−120sin3(θ)+144sin4(θ)=0
Usando el método de sustitución
−25−110sin2(θ)+120sin(θ)−120sin3(θ)+144sin4(θ)=0
Sea: sin(θ)=u−25−110u2+120u−120u3+144u4=0
−25−110u2+120u−120u3+144u4=0:u≈0.32943…,u≈0.88011…,u≈0.60823…,u≈−0.98445…
−25−110u2+120u−120u3+144u4=0
Escribir en la forma binómica an​xn+…+a1​x+a0​=0144u4−120u3−110u2+120u−25=0
Encontrar una solución para 144u4−120u3−110u2+120u−25=0 utilizando el método de Newton-Raphson:u≈0.32943…
144u4−120u3−110u2+120u−25=0
Definición del método de Newton-Raphson
f(u)=144u4−120u3−110u2+120u−25
Hallar f′(u):576u3−360u2−220u+120
dud​(144u4−120u3−110u2+120u−25)
Aplicar la regla de la suma/diferencia: (f±g)′=f′±g′=dud​(144u4)−dud​(120u3)−dud​(110u2)+dud​(120u)−dud​(25)
dud​(144u4)=576u3
dud​(144u4)
Sacar la constante: (a⋅f)′=a⋅f′=144dud​(u4)
Aplicar la regla de la potencia: dxd​(xa)=a⋅xa−1=144⋅4u4−1
Simplificar=576u3
dud​(120u3)=360u2
dud​(120u3)
Sacar la constante: (a⋅f)′=a⋅f′=120dud​(u3)
Aplicar la regla de la potencia: dxd​(xa)=a⋅xa−1=120⋅3u3−1
Simplificar=360u2
dud​(110u2)=220u
dud​(110u2)
Sacar la constante: (a⋅f)′=a⋅f′=110dud​(u2)
Aplicar la regla de la potencia: dxd​(xa)=a⋅xa−1=110⋅2u2−1
Simplificar=220u
dud​(120u)=120
dud​(120u)
Sacar la constante: (a⋅f)′=a⋅f′=120dudu​
Aplicar la regla de derivación: dudu​=1=120⋅1
Simplificar=120
dud​(25)=0
dud​(25)
Derivada de una constante: dxd​(a)=0=0
=576u3−360u2−220u+120−0
Simplificar=576u3−360u2−220u+120
Sea u0​=0Calcular un+1​ hasta que Δun+1​<0.000001
u1​=0.20833…:Δu1​=0.20833…
f(u0​)=144⋅04−120⋅03−110⋅02+120⋅0−25=−25f′(u0​)=576⋅03−360⋅02−220⋅0+120=120u1​=0.20833…
Δu1​=∣0.20833…−0∣=0.20833…Δu1​=0.20833…
u2​=0.29598…:Δu2​=0.08765…
f(u1​)=144⋅0.20833…4−120⋅0.20833…3−110⋅0.20833…2+120⋅0.20833…−25=−5.58810…f′(u1​)=576⋅0.20833…3−360⋅0.20833…2−220⋅0.20833…+120=63.75u2​=0.29598…
Δu2​=∣0.29598…−0.20833…∣=0.08765…Δu2​=0.08765…
u3​=0.32537…:Δu3​=0.02938…
f(u2​)=144⋅0.29598…4−120⋅0.29598…3−110⋅0.29598…2+120⋅0.29598…−25=−1.12484…f′(u2​)=576⋅0.29598…3−360⋅0.29598…2−220⋅0.29598…+120=38.27925…u3​=0.32537…
Δu3​=∣0.32537…−0.29598…∣=0.02938…Δu3​=0.02938…
u4​=0.32936…:Δu4​=0.00398…
f(u3​)=144⋅0.32537…4−120⋅0.32537…3−110⋅0.32537…2+120⋅0.32537…−25=−0.12024…f′(u3​)=576⋅0.32537…3−360⋅0.32537…2−220⋅0.32537…+120=30.14620…u4​=0.32936…
Δu4​=∣0.32936…−0.32537…∣=0.00398…Δu4​=0.00398…
u5​=0.32943…:Δu5​=0.00007…
f(u4​)=144⋅0.32936…4−120⋅0.32936…3−110⋅0.32936…2+120⋅0.32936…−25=−0.00215…f′(u4​)=576⋅0.32936…3−360⋅0.32936…2−220⋅0.32936…+120=29.06722…u5​=0.32943…
Δu5​=∣0.32943…−0.32936…∣=0.00007…Δu5​=0.00007…
u6​=0.32943…:Δu6​=2.54922E−8
f(u5​)=144⋅0.32943…4−120⋅0.32943…3−110⋅0.32943…2+120⋅0.32943…−25=−7.40478E−7f′(u5​)=576⋅0.32943…3−360⋅0.32943…2−220⋅0.32943…+120=29.04723…u6​=0.32943…
Δu6​=∣0.32943…−0.32943…∣=2.54922E−8Δu6​=2.54922E−8
u≈0.32943…
Aplicar la división larga Equation0:u−0.32943…144u4−120u3−110u2+120u−25​=144u3−72.56094…u2−133.90432…u+75.88684…
144u3−72.56094…u2−133.90432…u+75.88684…≈0
Encontrar una solución para 144u3−72.56094…u2−133.90432…u+75.88684…=0 utilizando el método de Newton-Raphson:u≈0.88011…
144u3−72.56094…u2−133.90432…u+75.88684…=0
Definición del método de Newton-Raphson
f(u)=144u3−72.56094…u2−133.90432…u+75.88684…
Hallar f′(u):432u2−145.12189…u−133.90432…
dud​(144u3−72.56094…u2−133.90432…u+75.88684…)
Aplicar la regla de la suma/diferencia: (f±g)′=f′±g′=dud​(144u3)−dud​(72.56094…u2)−dud​(133.90432…u)+dud​(75.88684…)
dud​(144u3)=432u2
dud​(144u3)
Sacar la constante: (a⋅f)′=a⋅f′=144dud​(u3)
Aplicar la regla de la potencia: dxd​(xa)=a⋅xa−1=144⋅3u3−1
Simplificar=432u2
dud​(72.56094…u2)=145.12189…u
dud​(72.56094…u2)
Sacar la constante: (a⋅f)′=a⋅f′=72.56094…dud​(u2)
Aplicar la regla de la potencia: dxd​(xa)=a⋅xa−1=72.56094…⋅2u2−1
Simplificar=145.12189…u
dud​(133.90432…u)=133.90432…
dud​(133.90432…u)
Sacar la constante: (a⋅f)′=a⋅f′=133.90432…dudu​
Aplicar la regla de derivación: dudu​=1=133.90432…⋅1
Simplificar=133.90432…
dud​(75.88684…)=0
dud​(75.88684…)
Derivada de una constante: dxd​(a)=0=0
=432u2−145.12189…u−133.90432…+0
Simplificar=432u2−145.12189…u−133.90432…
Sea u0​=1Calcular un+1​ hasta que Δun+1​<0.000001
u1​=0.91226…:Δu1​=0.08773…
f(u0​)=144⋅13−72.56094…⋅12−133.90432…⋅1+75.88684…=13.42157…f′(u0​)=432⋅12−145.12189…⋅1−133.90432…=152.97377…u1​=0.91226…
Δu1​=∣0.91226…−1∣=0.08773…Δu1​=0.08773…
u2​=0.88362…:Δu2​=0.02863…
f(u1​)=144⋅0.91226…3−72.56094…⋅0.91226…2−133.90432…⋅0.91226…+75.88684…=2.66967…f′(u1​)=432⋅0.91226…2−145.12189…⋅0.91226…−133.90432…=93.22653…u2​=0.88362…
Δu2​=∣0.88362…−0.91226…∣=0.02863…Δu2​=0.02863…
u3​=0.88016…:Δu3​=0.00346…
f(u2​)=144⋅0.88362…3−72.56094…⋅0.88362…2−133.90432…⋅0.88362…+75.88684…=0.26029…f′(u2​)=432⋅0.88362…2−145.12189…⋅0.88362…−133.90432…=75.16549…u3​=0.88016…
Δu3​=∣0.88016…−0.88362…∣=0.00346…Δu3​=0.00346…
u4​=0.88011…:Δu4​=0.00005…
f(u3​)=144⋅0.88016…3−72.56094…⋅0.88016…2−133.90432…⋅0.88016…+75.88684…=0.00370…f′(u3​)=432⋅0.88016…2−145.12189…⋅0.88016…−133.90432…=73.02944…u4​=0.88011…
Δu4​=∣0.88011…−0.88016…∣=0.00005…Δu4​=0.00005…
u5​=0.88011…:Δu5​=1.08272E−8
f(u4​)=144⋅0.88011…3−72.56094…⋅0.88011…2−133.90432…⋅0.88011…+75.88684…=7.90368E−7f′(u4​)=432⋅0.88011…2−145.12189…⋅0.88011…−133.90432…=72.99825…u5​=0.88011…
Δu5​=∣0.88011…−0.88011…∣=1.08272E−8Δu5​=1.08272E−8
u≈0.88011…
Aplicar la división larga Equation0:u−0.88011…144u3−72.56094…u2−133.90432…u+75.88684…​=144u2+54.17521…u−86.22405…
144u2+54.17521…u−86.22405…≈0
Encontrar una solución para 144u2+54.17521…u−86.22405…=0 utilizando el método de Newton-Raphson:u≈0.60823…
144u2+54.17521…u−86.22405…=0
Definición del método de Newton-Raphson
f(u)=144u2+54.17521…u−86.22405…
Hallar f′(u):288u+54.17521…
dud​(144u2+54.17521…u−86.22405…)
Aplicar la regla de la suma/diferencia: (f±g)′=f′±g′=dud​(144u2)+dud​(54.17521…u)−dud​(86.22405…)
dud​(144u2)=288u
dud​(144u2)
Sacar la constante: (a⋅f)′=a⋅f′=144dud​(u2)
Aplicar la regla de la potencia: dxd​(xa)=a⋅xa−1=144⋅2u2−1
Simplificar=288u
dud​(54.17521…u)=54.17521…
dud​(54.17521…u)
Sacar la constante: (a⋅f)′=a⋅f′=54.17521…dudu​
Aplicar la regla de derivación: dudu​=1=54.17521…⋅1
Simplificar=54.17521…
dud​(86.22405…)=0
dud​(86.22405…)
Derivada de una constante: dxd​(a)=0=0
=288u+54.17521…−0
Simplificar=288u+54.17521…
Sea u0​=2Calcular un+1​ hasta que Δun+1​<0.000001
u1​=1.05085…:Δu1​=0.94914…
f(u0​)=144⋅22+54.17521…⋅2−86.22405…=598.12636…f′(u0​)=288⋅2+54.17521…=630.17521…u1​=1.05085…
Δu1​=∣1.05085…−2∣=0.94914…Δu1​=0.94914…
u2​=0.68729…:Δu2​=0.36355…
f(u1​)=144⋅1.05085…2+54.17521…⋅1.05085…−86.22405…=129.72561…f′(u1​)=288⋅1.05085…+54.17521…=356.82203…u2​=0.68729…
Δu2​=∣0.68729…−1.05085…∣=0.36355…Δu2​=0.36355…
u3​=0.61180…:Δu3​=0.07549…
f(u2​)=144⋅0.68729…2+54.17521…⋅0.68729…−86.22405…=19.03314…f′(u2​)=288⋅0.68729…+54.17521…=252.11724…u3​=0.61180…
Δu3​=∣0.61180…−0.68729…∣=0.07549…Δu3​=0.07549…
u4​=0.60824…:Δu4​=0.00356…
f(u3​)=144⋅0.61180…2+54.17521…⋅0.61180…−86.22405…=0.82068…f′(u3​)=288⋅0.61180…+54.17521…=230.37518…u4​=0.60824…
Δu4​=∣0.60824…−0.61180…∣=0.00356…Δu4​=0.00356…
u5​=0.60823…:Δu5​=7.96803E−6
f(u4​)=144⋅0.60824…2+54.17521…⋅0.60824…−86.22405…=0.00182…f′(u4​)=288⋅0.60824…+54.17521…=229.34921…u5​=0.60823…
Δu5​=∣0.60823…−0.60824…∣=7.96803E−6Δu5​=7.96803E−6
u6​=0.60823…:Δu6​=3.98632E−11
f(u5​)=144⋅0.60823…2+54.17521…⋅0.60823…−86.22405…=9.1425E−9f′(u5​)=288⋅0.60823…+54.17521…=229.34692…u6​=0.60823…
Δu6​=∣0.60823…−0.60823…∣=3.98632E−11Δu6​=3.98632E−11
u≈0.60823…
Aplicar la división larga Equation0:u−0.60823…144u2+54.17521…u−86.22405…​=144u+141.76106…
144u+141.76106…≈0
u≈−0.98445…
Las soluciones sonu≈0.32943…,u≈0.88011…,u≈0.60823…,u≈−0.98445…
Sustituir en la ecuación u=sin(θ)sin(θ)≈0.32943…,sin(θ)≈0.88011…,sin(θ)≈0.60823…,sin(θ)≈−0.98445…
sin(θ)≈0.32943…,sin(θ)≈0.88011…,sin(θ)≈0.60823…,sin(θ)≈−0.98445…
sin(θ)=0.32943…:θ=arcsin(0.32943…)+2πn,θ=π−arcsin(0.32943…)+2πn
sin(θ)=0.32943…
Aplicar propiedades trigonométricas inversas
sin(θ)=0.32943…
Soluciones generales para sin(θ)=0.32943…sin(x)=a⇒x=arcsin(a)+2πn,x=π−arcsin(a)+2πnθ=arcsin(0.32943…)+2πn,θ=π−arcsin(0.32943…)+2πn
θ=arcsin(0.32943…)+2πn,θ=π−arcsin(0.32943…)+2πn
sin(θ)=0.88011…:θ=arcsin(0.88011…)+2πn,θ=π−arcsin(0.88011…)+2πn
sin(θ)=0.88011…
Aplicar propiedades trigonométricas inversas
sin(θ)=0.88011…
Soluciones generales para sin(θ)=0.88011…sin(x)=a⇒x=arcsin(a)+2πn,x=π−arcsin(a)+2πnθ=arcsin(0.88011…)+2πn,θ=π−arcsin(0.88011…)+2πn
θ=arcsin(0.88011…)+2πn,θ=π−arcsin(0.88011…)+2πn
sin(θ)=0.60823…:θ=arcsin(0.60823…)+2πn,θ=π−arcsin(0.60823…)+2πn
sin(θ)=0.60823…
Aplicar propiedades trigonométricas inversas
sin(θ)=0.60823…
Soluciones generales para sin(θ)=0.60823…sin(x)=a⇒x=arcsin(a)+2πn,x=π−arcsin(a)+2πnθ=arcsin(0.60823…)+2πn,θ=π−arcsin(0.60823…)+2πn
θ=arcsin(0.60823…)+2πn,θ=π−arcsin(0.60823…)+2πn
sin(θ)=−0.98445…:θ=arcsin(−0.98445…)+2πn,θ=π+arcsin(0.98445…)+2πn
sin(θ)=−0.98445…
Aplicar propiedades trigonométricas inversas
sin(θ)=−0.98445…
Soluciones generales para sin(θ)=−0.98445…sin(x)=−a⇒x=arcsin(−a)+2πn,x=π+arcsin(a)+2πnθ=arcsin(−0.98445…)+2πn,θ=π+arcsin(0.98445…)+2πn
θ=arcsin(−0.98445…)+2πn,θ=π+arcsin(0.98445…)+2πn
Combinar toda las solucionesθ=arcsin(0.32943…)+2πn,θ=π−arcsin(0.32943…)+2πn,θ=arcsin(0.88011…)+2πn,θ=π−arcsin(0.88011…)+2πn,θ=arcsin(0.60823…)+2πn,θ=π−arcsin(0.60823…)+2πn,θ=arcsin(−0.98445…)+2πn,θ=π+arcsin(0.98445…)+2πn
Verificar las soluciones sustituyendo en la ecuación original
Verificar las soluciones sustituyéndolas en 3sec(θ)+5csc(θ)=12
Quitar las que no concuerden con la ecuación.
Verificar la solución arcsin(0.32943…)+2πn:Falso
arcsin(0.32943…)+2πn
Sustituir n=1arcsin(0.32943…)+2π1
Multiplicar 3sec(θ)+5csc(θ)=12 por θ=arcsin(0.32943…)+2π13sec(arcsin(0.32943…)+2π1)+5csc(arcsin(0.32943…)+2π1)=12
Simplificar18.35473…=12
⇒Falso
Verificar la solución π−arcsin(0.32943…)+2πn:Verdadero
π−arcsin(0.32943…)+2πn
Sustituir n=1π−arcsin(0.32943…)+2π1
Multiplicar 3sec(θ)+5csc(θ)=12 por θ=π−arcsin(0.32943…)+2π13sec(π−arcsin(0.32943…)+2π1)+5csc(π−arcsin(0.32943…)+2π1)=12
Simplificar12=12
⇒Verdadero
Verificar la solución arcsin(0.88011…)+2πn:Verdadero
arcsin(0.88011…)+2πn
Sustituir n=1arcsin(0.88011…)+2π1
Multiplicar 3sec(θ)+5csc(θ)=12 por θ=arcsin(0.88011…)+2π13sec(arcsin(0.88011…)+2π1)+5csc(arcsin(0.88011…)+2π1)=12
Simplificar12=12
⇒Verdadero
Verificar la solución π−arcsin(0.88011…)+2πn:Falso
π−arcsin(0.88011…)+2πn
Sustituir n=1π−arcsin(0.88011…)+2π1
Multiplicar 3sec(θ)+5csc(θ)=12 por θ=π−arcsin(0.88011…)+2π13sec(π−arcsin(0.88011…)+2π1)+5csc(π−arcsin(0.88011…)+2π1)=12
Simplificar−0.63781…=12
⇒Falso
Verificar la solución arcsin(0.60823…)+2πn:Verdadero
arcsin(0.60823…)+2πn
Sustituir n=1arcsin(0.60823…)+2π1
Multiplicar 3sec(θ)+5csc(θ)=12 por θ=arcsin(0.60823…)+2π13sec(arcsin(0.60823…)+2π1)+5csc(arcsin(0.60823…)+2π1)=12
Simplificar12=12
⇒Verdadero
Verificar la solución π−arcsin(0.60823…)+2πn:Falso
π−arcsin(0.60823…)+2πn
Sustituir n=1π−arcsin(0.60823…)+2π1
Multiplicar 3sec(θ)+5csc(θ)=12 por θ=π−arcsin(0.60823…)+2π13sec(π−arcsin(0.60823…)+2π1)+5csc(π−arcsin(0.60823…)+2π1)=12
Simplificar4.44101…=12
⇒Falso
Verificar la solución arcsin(−0.98445…)+2πn:Verdadero
arcsin(−0.98445…)+2πn
Sustituir n=1arcsin(−0.98445…)+2π1
Multiplicar 3sec(θ)+5csc(θ)=12 por θ=arcsin(−0.98445…)+2π13sec(arcsin(−0.98445…)+2π1)+5csc(arcsin(−0.98445…)+2π1)=12
Simplificar12=12
⇒Verdadero
Verificar la solución π+arcsin(0.98445…)+2πn:Falso
π+arcsin(0.98445…)+2πn
Sustituir n=1π+arcsin(0.98445…)+2π1
Multiplicar 3sec(θ)+5csc(θ)=12 por θ=π+arcsin(0.98445…)+2π13sec(π+arcsin(0.98445…)+2π1)+5csc(π+arcsin(0.98445…)+2π1)=12
Simplificar−22.15793…=12
⇒Falso
θ=π−arcsin(0.32943…)+2πn,θ=arcsin(0.88011…)+2πn,θ=arcsin(0.60823…)+2πn,θ=arcsin(−0.98445…)+2πn
Mostrar soluciones en forma decimalθ=π−0.33570…+2πn,θ=1.07609…+2πn,θ=0.65383…+2πn,θ=−1.39422…+2πn

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