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tan^4(x)+tan^2(x)=tan(x)

Solución

tan^{4} (x) + tan^{2} (x) = tan (x)

Solución

x = πn, x = 0.59876 \dots + πn

Grados

x = 0^{\circ} + 18 0^{\circ} n, x = 34.30680 \dots^{\circ} + 18 0^{\circ} n

Pasos de solución

tan^{4} (x) + tan^{2} (x) = tan (x)

Usando el método de sustitución

tan^{4} (x) + tan^{2} (x) = tan (x)

Sea:

tan (x) = u

u^{4} + u^{2} = u

u^{4} + u^{2} = u : u = 0, u \approx 0.68232 \dots

u^{4} + u^{2} = u

Mover

u

al lado izquierdo

u^{4} + u^{2} = u

Restar

u

de ambos lados

u^{4} + u^{2} - u = u - u

Simplificar

u^{4} + u^{2} - u = 0

u^{4} + u^{2} - u = 0

Factorizar

u^{4} + u^{2} - u : u (u^{3} + u - 1)

u^{4} + u^{2} - u

Aplicar las leyes de los exponentes:

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

u^{2} = uu

= u^{3} u + uu - u

Factorizar el termino común

u

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

u (u^{3} + u - 1) = 0

Utilizando el teorema de factor cero: si

ab = 0

entonces

a = 0

b = 0

u = 0 or u^{3} + u - 1 = 0

Resolver

u^{3} + u - 1 = 0 : u \approx 0.68232 \dots

u^{3} + u - 1 = 0

Encontrar una solución para

u^{3} + u - 1 = 0

utilizando el método de Newton-Raphson:

u \approx 0.68232 \dots

u^{3} + u - 1 = 0

Definición del método de Newton-Raphson

f (u) = u^{3} + u - 1

Hallar

f^{^{'}} (u) : 3 u^{2} + 1

\frac{d}{d u} (u^{3} + u - 1)

Aplicar la regla de la suma/diferencia:

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

= \frac{d}{d u} (u^{3}) + \frac{d u}{d u} - \frac{d}{d u} (1)

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

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

Aplicar la regla de la potencia:

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

= 3 u^{3 - 1}

Simplificar

= 3 u^{2}

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

\frac{d u}{d u}

Aplicar la regla de derivación:

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

= 1

\frac{d}{d u} (1) = 0

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

Derivada de una constante:

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

= 0

= 3 u^{2} + 1 - 0

Simplificar

= 3 u^{2} + 1

Sea

u_{0} = 1

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = 0.75 : Δ u_{1} = 0.25

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

f^{^{'}} (u_{0}) = 3 \cdot 1^{2} + 1 = 4

u_{1} = 0.75

Δ u_{1} = ∣ 0.75 - 1 ∣ = 0.25

Δ u_{1} = 0.25

u_{2} = 0.68604 \dots : Δ u_{2} = 0.06395 \dots

f (u_{1}) = 0.7 5^{3} + 0.75 - 1 = 0.171875

f^{^{'}} (u_{1}) = 3 \cdot 0.7 5^{2} + 1 = 2.6875

u_{2} = 0.68604 \dots

Δ u_{2} = ∣ 0.68604 \dots - 0.75 ∣ = 0.06395 \dots

Δ u_{2} = 0.06395 \dots

u_{3} = 0.68233 \dots : Δ u_{3} = 0.00370 \dots

f (u_{2}) = 0.68604 \dots^{3} + 0.68604 \dots - 1 = 0.00894 \dots

f^{^{'}} (u_{2}) = 3 \cdot 0.68604 \dots^{2} + 1 = 2.41197 \dots

u_{3} = 0.68233 \dots

Δ u_{3} = ∣ 0.68233 \dots - 0.68604 \dots ∣ = 0.00370 \dots

Δ u_{3} = 0.00370 \dots

u_{4} = 0.68232 \dots : Δ u_{4} = 0.00001 \dots

f (u_{3}) = 0.68233 \dots^{3} + 0.68233 \dots - 1 = 0.00002 \dots

f^{^{'}} (u_{3}) = 3 \cdot 0.68233 \dots^{2} + 1 = 2.39676 \dots

u_{4} = 0.68232 \dots

Δ u_{4} = ∣ 0.68232 \dots - 0.68233 \dots ∣ = 0.00001 \dots

Δ u_{4} = 0.00001 \dots

u_{5} = 0.68232 \dots : Δ u_{5} = 1.18493 E - 10

f (u_{4}) = 0.68232 \dots^{3} + 0.68232 \dots - 1 = 2.83995 E - 10

f^{^{'}} (u_{4}) = 3 \cdot 0.68232 \dots^{2} + 1 = 2.39671 \dots

u_{5} = 0.68232 \dots

Δ u_{5} = ∣ 0.68232 \dots - 0.68232 \dots ∣ = 1.18493 E - 10

Δ u_{5} = 1.18493 E - 10

u \approx 0.68232 \dots

Aplicar la división larga

Eq u a t i o n 0 : \frac{u ^{3} + u - 1}{u - 0.68232 \dots} = u^{2} + 0.68232 \dots u + 1.46557 \dots

u^{2} + 0.68232 \dots u + 1.46557 \dots \approx 0

Encontrar una solución para

u^{2} + 0.68232 \dots u + 1.46557 \dots = 0

utilizando el método de Newton-Raphson:

Sin solución para

u \in R

u^{2} + 0.68232 \dots u + 1.46557 \dots = 0

Definición del método de Newton-Raphson

f (u) = u^{2} + 0.68232 \dots u + 1.46557 \dots

Hallar

f^{^{'}} (u) : 2 u + 0.68232 \dots

\frac{d}{d u} (u^{2} + 0.68232 \dots u + 1.46557 \dots)

Aplicar la regla de la suma/diferencia:

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

= \frac{d}{d u} (u^{2}) + \frac{d}{d u} (0.68232 \dots u) + \frac{d}{d u} (1.46557 \dots)

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

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

Aplicar la regla de la potencia:

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

= 2 u^{2 - 1}

Simplificar

= 2 u

\frac{d}{d u} (0.68232 \dots u) = 0.68232 \dots

\frac{d}{d u} (0.68232 \dots u)

Sacar la constante:

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

= 0.68232 \dots \frac{d u}{d u}

Aplicar la regla de derivación:

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

= 0.68232 \dots \cdot 1

Simplificar

= 0.68232 \dots

\frac{d}{d u} (1.46557 \dots) = 0

\frac{d}{d u} (1.46557 \dots)

Derivada de una constante:

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

= 0

= 2 u + 0.68232 \dots + 0

Simplificar

= 2 u + 0.68232 \dots

Sea

u_{0} = - 2

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = - 0.76391 \dots : Δ u_{1} = 1.23608 \dots

f (u_{0}) = (- 2)^{2} + 0.68232 \dots (- 2) + 1.46557 \dots = 4.10091 \dots

f^{^{'}} (u_{0}) = 2 (- 2) + 0.68232 \dots = - 3.31767 \dots

u_{1} = - 0.76391 \dots

Δ u_{1} = ∣ - 0.76391 \dots - (- 2) ∣ = 1.23608 \dots

Δ u_{1} = 1.23608 \dots

u_{2} = 1.04316 \dots : Δ u_{2} = 1.80707 \dots

f (u_{1}) = (- 0.76391 \dots)^{2} + 0.68232 \dots (- 0.76391 \dots) + 1.46557 \dots = 1.52789 \dots

f^{^{'}} (u_{1}) = 2 (- 0.76391 \dots) + 0.68232 \dots = - 0.84550 \dots

u_{2} = 1.04316 \dots

Δ u_{2} = ∣ 1.04316 \dots - (- 0.76391 \dots) ∣ = 1.80707 \dots

Δ u_{2} = 1.80707 \dots

u_{3} = - 0.13630 \dots : Δ u_{3} = 1.17946 \dots

f (u_{2}) = 1.04316 \dots^{2} + 0.68232 \dots \cdot 1.04316 \dots + 1.46557 \dots = 3.26553 \dots

f^{^{'}} (u_{2}) = 2 \cdot 1.04316 \dots + 0.68232 \dots = 2.76865 \dots

u_{3} = - 0.13630 \dots

Δ u_{3} = ∣ - 0.13630 \dots - 1.04316 \dots ∣ = 1.17946 \dots

Δ u_{3} = 1.17946 \dots

u_{4} = - 3.53171 \dots : Δ u_{4} = 3.39540 \dots

f (u_{3}) = (- 0.13630 \dots)^{2} + 0.68232 \dots (- 0.13630 \dots) + 1.46557 \dots = 1.39114 \dots

f^{^{'}} (u_{3}) = 2 (- 0.13630 \dots) + 0.68232 \dots = 0.40971 \dots

u_{4} = - 3.53171 \dots

Δ u_{4} = ∣ - 3.53171 \dots - (- 0.13630 \dots) ∣ = 3.39540 \dots

Δ u_{4} = 3.39540 \dots

u_{5} = - 1.72500 \dots : Δ u_{5} = 1.80670 \dots

f (u_{4}) = (- 3.53171 \dots)^{2} + 0.68232 \dots (- 3.53171 \dots) + 1.46557 \dots = 11.52876 \dots

f^{^{'}} (u_{4}) = 2 (- 3.53171 \dots) + 0.68232 \dots = - 6.38109 \dots

u_{5} = - 1.72500 \dots

Δ u_{5} = ∣ - 1.72500 \dots - (- 3.53171 \dots) ∣ = 1.80670 \dots

Δ u_{5} = 1.80670 \dots

u_{6} = - 0.54560 \dots : Δ u_{6} = 1.17939 \dots

f (u_{5}) = (- 1.72500 \dots)^{2} + 0.68232 \dots (- 1.72500 \dots) + 1.46557 \dots = 3.26419 \dots

f^{^{'}} (u_{5}) = 2 (- 1.72500 \dots) + 0.68232 \dots = - 2.76767 \dots

u_{6} = - 0.54560 \dots

Δ u_{6} = ∣ - 0.54560 \dots - (- 1.72500 \dots) ∣ = 1.17939 \dots

Δ u_{6} = 1.17939 \dots

u_{7} = 2.85625 \dots : Δ u_{7} = 3.40185 \dots

f (u_{6}) = (- 0.54560 \dots)^{2} + 0.68232 \dots (- 0.54560 \dots) + 1.46557 \dots = 1.39097 \dots

f^{^{'}} (u_{6}) = 2 (- 0.54560 \dots) + 0.68232 \dots = - 0.40888 \dots

u_{7} = 2.85625 \dots

Δ u_{7} = ∣ 2.85625 \dots - (- 0.54560 \dots) ∣ = 3.40185 \dots

Δ u_{7} = 3.40185 \dots

u_{8} = 1.04656 \dots : Δ u_{8} = 1.80968 \dots

f (u_{7}) = 2.85625 \dots^{2} + 0.68232 \dots \cdot 2.85625 \dots + 1.46557 \dots = 11.57264 \dots

f^{^{'}} (u_{7}) = 2 \cdot 2.85625 \dots + 0.68232 \dots = 6.39483 \dots

u_{8} = 1.04656 \dots

Δ u_{8} = ∣ 1.04656 \dots - 2.85625 \dots ∣ = 1.80968 \dots

Δ u_{8} = 1.80968 \dots

u_{9} = - 0.13340 \dots : Δ u_{9} = 1.17997 \dots

f (u_{8}) = 1.04656 \dots^{2} + 0.68232 \dots \cdot 1.04656 \dots + 1.46557 \dots = 3.27496 \dots

f^{^{'}} (u_{8}) = 2 \cdot 1.04656 \dots + 0.68232 \dots = 2.77545 \dots

u_{9} = - 0.13340 \dots

Δ u_{9} = ∣ - 0.13340 \dots - 1.04656 \dots ∣ = 1.17997 \dots

Δ u_{9} = 1.17997 \dots

u_{10} = - 3.48434 \dots : Δ u_{10} = 3.35093 \dots

f (u_{9}) = (- 0.13340 \dots)^{2} + 0.68232 \dots (- 0.13340 \dots) + 1.46557 \dots = 1.39234 \dots

f^{^{'}} (u_{9}) = 2 (- 0.13340 \dots) + 0.68232 \dots = 0.41550 \dots

u_{10} = - 3.48434 \dots

Δ u_{10} = ∣ - 3.48434 \dots - (- 0.13340 \dots) ∣ = 3.35093 \dots

Δ u_{10} = 3.35093 \dots

No se puede encontrar solución

La solución es

u \approx 0.68232 \dots

Las soluciones son

u = 0, u \approx 0.68232 \dots

Sustituir en la ecuación

u = tan (x)

tan (x) = 0, tan (x) \approx 0.68232 \dots

tan (x) = 0, tan (x) \approx 0.68232 \dots

tan (x) = 0 : x = πn

tan (x) = 0

Soluciones generales para

tan (x) = 0

tan (x)

tabla de valores periódicos con

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = 0 + πn

x = 0 + πn

Resolver

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = πn

tan (x) = 0.68232 \dots : x = arctan (0.68232 \dots) + πn

tan (x) = 0.68232 \dots

Aplicar propiedades trigonométricas inversas

tan (x) = 0.68232 \dots

Soluciones generales para

tan (x) = 0.68232 \dots

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

x = arctan (0.68232 \dots) + πn

x = arctan (0.68232 \dots) + πn

Combinar toda las soluciones

x = πn, x = arctan (0.68232 \dots) + πn

Mostrar soluciones en forma decimal

x = πn, x = 0.59876 \dots + πn

Ejemplos populares

sin^4(x)+sin^2(x)=sin^6(x)

sin^{4} (x) + sin^{2} (x) = sin^{6} (x)

cos(a)=(-11)/(14)

cos (a) = \frac{- 11}{14}

solvefor x,sin(x/x)=0.7 resolver para

x, sin (\frac{x}{x}) = 0.7

5cos^2(x)+sin^2(x)=4

5 cos^{2} (x) + sin^{2} (x) = 4

cos(u)-1.5sin^2(u)+0.1667=0

cos (u) - 1.5 sin^{2} (u) + 0.1667 = 0