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Popular Trigonometría >

cos^2(a)= 3/(5sin(a))

Solución

cos^{2} (a) = \frac{3}{5 sin ( a )}

Solución

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

Pasos de solución

cos^{2} (a) = \frac{3}{5 sin ( a )}

Restar

\frac{3}{5 sin ( a )}

de ambos lados

cos^{2} (a) - \frac{3}{5 sin ( a )} = 0

Simplificar

cos^{2} (a) - \frac{3}{5 sin ( a )} : \frac{5 cos ^{2} ( a ) sin ( a ) - 3}{5 sin ( a )}

cos^{2} (a) - \frac{3}{5 sin ( a )}

Convertir a fracción:

cos^{2} (a) = \frac{cos ^{2} ( a ) 5 sin ( a )}{5 sin ( a )}

= \frac{cos ^{2} ( a ) \cdot 5 sin ( a )}{5 sin ( a )} - \frac{3}{5 sin ( a )}

Ya que los denominadores son iguales, combinar las fracciones:

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

= \frac{cos ^{2} ( a ) \cdot 5 sin ( a ) - 3}{5 sin ( a )}

\frac{5 cos ^{2} ( a ) sin ( a ) - 3}{5 sin ( a )} = 0

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

5 cos^{2} (a) sin (a) - 3 = 0

Re-escribir usando identidades trigonométricas

- 3 + 5 cos^{2} (a) sin (a)

Utilizar la identidad pitagórica:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= - 3 + 5 (1 - sin^{2} (a)) sin (a)

- 3 + (1 - sin^{2} (a)) \cdot 5 sin (a) = 0

Usando el método de sustitución

- 3 + (1 - sin^{2} (a)) \cdot 5 sin (a) = 0

Sea:

sin (a) = u

- 3 + (1 - u^{2}) \cdot 5 u = 0

- 3 + (1 - u^{2}) \cdot 5 u = 0 : u \approx - 1.22119 \dots

- 3 + (1 - u^{2}) \cdot 5 u = 0

Desarrollar

- 3 + (1 - u^{2}) \cdot 5 u : - 3 + 5 u - 5 u^{3}

- 3 + (1 - u^{2}) \cdot 5 u

= - 3 + 5 u (1 - u^{2})

Expandir

5 u (1 - u^{2}) : 5 u - 5 u^{3}

5 u (1 - u^{2})

Poner los parentesis utilizando:

a (b - c) = ab - a c

a = 5 u, b = 1, c = u^{2}

= 5 u \cdot 1 - 5 u u^{2}

= 5 \cdot 1 \cdot u - 5 u^{2} u

Simplificar

5 \cdot 1 \cdot u - 5 u^{2} u : 5 u - 5 u^{3}

5 \cdot 1 \cdot u - 5 u^{2} u

5 \cdot 1 \cdot u = 5 u

5 \cdot 1 \cdot u

Multiplicar los numeros:

5 \cdot 1 = 5

= 5 u

5 u^{2} u = 5 u^{3}

5 u^{2} u

Aplicar las leyes de los exponentes:

a^{b} \cdot a^{c} = a^{b + c}

u^{2} u = u^{2 + 1}

= 5 u^{2 + 1}

Sumar:

2 + 1 = 3

= 5 u^{3}

= 5 u - 5 u^{3}

= 5 u - 5 u^{3}

= - 3 + 5 u - 5 u^{3}

- 3 + 5 u - 5 u^{3} = 0

Escribir en la forma binómica

a_{n} x^{n} + \dots + a_{1} x + a = 0

- 5 u^{3} + 5 u - 3 = 0

Encontrar una solución para

- 5 u^{3} + 5 u - 3 = 0

utilizando el método de Newton-Raphson:

u \approx - 1.22119 \dots

- 5 u^{3} + 5 u - 3 = 0

Definición del método de Newton-Raphson

f (u) = - 5 u^{3} + 5 u - 3

Hallar

f^{^{'}} (u) : - 15 u^{2} + 5

\frac{d}{d u} (- 5 u^{3} + 5 u - 3)

Aplicar la regla de la suma/diferencia:

(f \pm g)^{'} = f^{'} \pm g^{'}

= - \frac{d}{d u} (5 u^{3}) + \frac{d}{d u} (5 u) - \frac{d}{d u} (3)

\frac{d}{d u} (5 u^{3}) = 15 u^{2}

\frac{d}{d u} (5 u^{3})

Sacar la constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 5 \frac{d}{d u} (u^{3})

Aplicar la regla de la potencia:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 5 \cdot 3 u^{3 - 1}

Simplificar

= 15 u^{2}

\frac{d}{d u} (5 u) = 5

\frac{d}{d u} (5 u)

Sacar la constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 5 \frac{d u}{d u}

Aplicar la regla de derivación:

\frac{d u}{d u} = 1

= 5 \cdot 1

Simplificar

= 5

\frac{d}{d u} (3) = 0

\frac{d}{d u} (3)

Derivada de una constante:

\frac{d}{d x} (a) = 0

= 0

= - 15 u^{2} + 5 - 0

Simplificar

= - 15 u^{2} + 5

Sea

u_{0} = - 1

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = - 1.3 : Δ u_{1} = 0.3

f (u_{0}) = - 5 (- 1)^{3} + 5 (- 1) - 3 = - 3

f^{^{'}} (u_{0}) = - 15 (- 1)^{2} + 5 = - 10

u_{1} = - 1.3

Δ u_{1} = ∣ - 1.3 - (- 1) ∣ = 0.3

Δ u_{1} = 0.3

u_{2} = - 1.22702 \dots : Δ u_{2} = 0.07297 \dots

f (u_{1}) = - 5 (- 1.3)^{3} + 5 (- 1.3) - 3 = 1.485

f^{^{'}} (u_{1}) = - 15 (- 1.3)^{2} + 5 = - 20.35

u_{2} = - 1.22702 \dots

Δ u_{2} = ∣ - 1.22702 \dots - (- 1.3) ∣ = 0.07297 \dots

Δ u_{2} = 0.07297 \dots

u_{3} = - 1.22123 \dots : Δ u_{3} = 0.00579 \dots

f (u_{2}) = - 5 (- 1.22702 \dots)^{3} + 5 (- 1.22702 \dots) - 3 = 0.10189 \dots

f^{^{'}} (u_{2}) = - 15 (- 1.22702 \dots)^{2} + 5 = - 17.58392 \dots

u_{3} = - 1.22123 \dots

Δ u_{3} = ∣ - 1.22123 \dots - (- 1.22702 \dots) ∣ = 0.00579 \dots

Δ u_{3} = 0.00579 \dots

u_{4} = - 1.22119 \dots : Δ u_{4} = 0.00003 \dots

f (u_{3}) = - 5 (- 1.22123 \dots)^{3} + 5 (- 1.22123 \dots) - 3 = 0.00061 \dots

f^{^{'}} (u_{3}) = - 15 (- 1.22123 \dots)^{2} + 5 = - 17.37112 \dots

u_{4} = - 1.22119 \dots

Δ u_{4} = ∣ - 1.22119 \dots - (- 1.22123 \dots) ∣ = 0.00003 \dots

Δ u_{4} = 0.00003 \dots

u_{5} = - 1.22119 \dots : Δ u_{5} = 1.33081 E - 9

f (u_{4}) = - 5 (- 1.22119 \dots)^{3} + 5 (- 1.22119 \dots) - 3 = 2.31159 E - 8

f^{^{'}} (u_{4}) = - 15 (- 1.22119 \dots)^{2} + 5 = - 17.36982 \dots

u_{5} = - 1.22119 \dots

Δ u_{5} = ∣ - 1.22119 \dots - (- 1.22119 \dots) ∣ = 1.33081 E - 9

Δ u_{5} = 1.33081 E - 9

u \approx - 1.22119 \dots

Aplicar la división larga

Eq u a t i o n 0 : \frac{- 5 u ^{3} + 5 u - 3}{u + 1.22119 \dots} = - 5 u^{2} + 6.10598 \dots u - 2.45660 \dots

- 5 u^{2} + 6.10598 \dots u - 2.45660 \dots \approx 0

Encontrar una solución para

- 5 u^{2} + 6.10598 \dots u - 2.45660 \dots = 0

utilizando el método de Newton-Raphson:

Sin solución para

u \in R

- 5 u^{2} + 6.10598 \dots u - 2.45660 \dots = 0

Definición del método de Newton-Raphson

f (u) = - 5 u^{2} + 6.10598 \dots u - 2.45660 \dots

Hallar

f^{^{'}} (u) : - 10 u + 6.10598 \dots

\frac{d}{d u} (- 5 u^{2} + 6.10598 \dots u - 2.45660 \dots)

Aplicar la regla de la suma/diferencia:

(f \pm g)^{'} = f^{'} \pm g^{'}

= - \frac{d}{d u} (5 u^{2}) + \frac{d}{d u} (6.10598 \dots u) - \frac{d}{d u} (2.45660 \dots)

\frac{d}{d u} (5 u^{2}) = 10 u

\frac{d}{d u} (5 u^{2})

Sacar la constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 5 \frac{d}{d u} (u^{2})

Aplicar la regla de la potencia:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 5 \cdot 2 u^{2 - 1}

Simplificar

= 10 u

\frac{d}{d u} (6.10598 \dots u) = 6.10598 \dots

\frac{d}{d u} (6.10598 \dots u)

Sacar la constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 6.10598 \dots \frac{d u}{d u}

Aplicar la regla de derivación:

\frac{d u}{d u} = 1

= 6.10598 \dots \cdot 1

Simplificar

= 6.10598 \dots

\frac{d}{d u} (2.45660 \dots) = 0

\frac{d}{d u} (2.45660 \dots)

Derivada de una constante:

\frac{d}{d x} (a) = 0

= 0

= - 10 u + 6.10598 \dots - 0

Simplificar

= - 10 u + 6.10598 \dots

Sea

u_{0} = 0

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = 0.40232 \dots : Δ u_{1} = 0.40232 \dots

f (u_{0}) = - 5 \cdot 0^{2} + 6.10598 \dots \cdot 0 - 2.45660 \dots = - 2.45660 \dots

f^{^{'}} (u_{0}) = - 10 \cdot 0 + 6.10598 \dots = 6.10598 \dots

u_{1} = 0.40232 \dots

Δ u_{1} = ∣ 0.40232 \dots - 0 ∣ = 0.40232 \dots

Δ u_{1} = 0.40232 \dots

u_{2} = 0.79092 \dots : Δ u_{2} = 0.38859 \dots

f (u_{1}) = - 5 \cdot 0.40232 \dots^{2} + 6.10598 \dots \cdot 0.40232 \dots - 2.45660 \dots = - 0.80933 \dots

f^{^{'}} (u_{1}) = - 10 \cdot 0.40232 \dots + 6.10598 \dots = 2.08270 \dots

u_{2} = 0.79092 \dots

Δ u_{2} = ∣ 0.79092 \dots - 0.40232 \dots ∣ = 0.38859 \dots

Δ u_{2} = 0.38859 \dots

u_{3} = 0.37222 \dots : Δ u_{3} = 0.41870 \dots

f (u_{2}) = - 5 \cdot 0.79092 \dots^{2} + 6.10598 \dots \cdot 0.79092 \dots - 2.45660 \dots = - 0.75504 \dots

f^{^{'}} (u_{2}) = - 10 \cdot 0.79092 \dots + 6.10598 \dots = - 1.80328 \dots

u_{3} = 0.37222 \dots

Δ u_{3} = ∣ 0.37222 \dots - 0.79092 \dots ∣ = 0.41870 \dots

Δ u_{3} = 0.41870 \dots

u_{4} = 0.73994 \dots : Δ u_{4} = 0.36772 \dots

f (u_{3}) = - 5 \cdot 0.37222 \dots^{2} + 6.10598 \dots \cdot 0.37222 \dots - 2.45660 \dots = - 0.87657 \dots

f^{^{'}} (u_{3}) = - 10 \cdot 0.37222 \dots + 6.10598 \dots = 2.38377 \dots

u_{4} = 0.73994 \dots

Δ u_{4} = ∣ 0.73994 \dots - 0.37222 \dots ∣ = 0.36772 \dots

Δ u_{4} = 0.36772 \dots

u_{5} = 0.21724 \dots : Δ u_{5} = 0.52270 \dots

f (u_{4}) = - 5 \cdot 0.73994 \dots^{2} + 6.10598 \dots \cdot 0.73994 \dots - 2.45660 \dots = - 0.67611 \dots

f^{^{'}} (u_{4}) = - 10 \cdot 0.73994 \dots + 6.10598 \dots = - 1.29348 \dots

u_{5} = 0.21724 \dots

Δ u_{5} = ∣ 0.21724 \dots - 0.73994 \dots ∣ = 0.52270 \dots

Δ u_{5} = 0.52270 \dots

u_{6} = 0.56453 \dots : Δ u_{6} = 0.34729 \dots

f (u_{5}) = - 5 \cdot 0.21724 \dots^{2} + 6.10598 \dots \cdot 0.21724 \dots - 2.45660 \dots = - 1.36610 \dots

f^{^{'}} (u_{5}) = - 10 \cdot 0.21724 \dots + 6.10598 \dots = 3.93356 \dots

u_{6} = 0.56453 \dots

Δ u_{6} = ∣ 0.56453 \dots - 0.21724 \dots ∣ = 0.34729 \dots

Δ u_{6} = 0.34729 \dots

u_{7} = 1.87375 \dots : Δ u_{7} = 1.30922 \dots

f (u_{6}) = - 5 \cdot 0.56453 \dots^{2} + 6.10598 \dots \cdot 0.56453 \dots - 2.45660 \dots = - 0.60306 \dots

f^{^{'}} (u_{6}) = - 10 \cdot 0.56453 \dots + 6.10598 \dots = 0.46062 \dots

u_{7} = 1.87375 \dots

Δ u_{7} = ∣ 1.87375 \dots - 0.56453 \dots ∣ = 1.30922 \dots

Δ u_{7} = 1.30922 \dots

u_{8} = 1.19527 \dots : Δ u_{8} = 0.67848 \dots

f (u_{7}) = - 5 \cdot 1.87375 \dots^{2} + 6.10598 \dots \cdot 1.87375 \dots - 2.45660 \dots = - 8.57033 \dots

f^{^{'}} (u_{7}) = - 10 \cdot 1.87375 \dots + 6.10598 \dots = - 12.63161 \dots

u_{8} = 1.19527 \dots

Δ u_{8} = ∣ 1.19527 \dots - 1.87375 \dots ∣ = 0.67848 \dots

Δ u_{8} = 0.67848 \dots

u_{9} = 0.80160 \dots : Δ u_{9} = 0.39366 \dots

f (u_{8}) = - 5 \cdot 1.19527 \dots^{2} + 6.10598 \dots \cdot 1.19527 \dots - 2.45660 \dots = - 2.30169 \dots

f^{^{'}} (u_{8}) = - 10 \cdot 1.19527 \dots + 6.10598 \dots = - 5.84678 \dots

u_{9} = 0.80160 \dots

Δ u_{9} = ∣ 0.80160 \dots - 1.19527 \dots ∣ = 0.39366 \dots

Δ u_{9} = 0.39366 \dots

u_{10} = 0.39593 \dots : Δ u_{10} = 0.40567 \dots

f (u_{9}) = - 5 \cdot 0.80160 \dots^{2} + 6.10598 \dots \cdot 0.80160 \dots - 2.45660 \dots = - 0.77487 \dots

f^{^{'}} (u_{9}) = - 10 \cdot 0.80160 \dots + 6.10598 \dots = - 1.91008 \dots

u_{10} = 0.39593 \dots

Δ u_{10} = ∣ 0.39593 \dots - 0.80160 \dots ∣ = 0.40567 \dots

Δ u_{10} = 0.40567 \dots

No se puede encontrar solución

La solución es

u \approx - 1.22119 \dots

Sustituir en la ecuación

u = sin (a)

sin (a) \approx - 1.22119 \dots

sin (a) \approx - 1.22119 \dots

sin (a) = - 1.22119 \dots :

Sin solución

sin (a) = - 1.22119 \dots

- 1 \leq sin (x) \leq 1

Sin soluci \overset{o}{ˊ} n

Combinar toda las soluciones

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

Gráfica

Ejemplos populares

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2+cos^2(x)=5sin(x)