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tan^2(a)=((2tan(a)))/((1-tan^2(a)))

Solución

tan^{2} (a) = \frac{( 2 tan ( a ) )}{( 1 - tan ^{2} ( a ) )}

Solución

a = πn, a = - 0.98930 \dots + πn

Grados

a = 0^{\circ} + 18 0^{\circ} n, a = - 56.68315 \dots^{\circ} + 18 0^{\circ} n

Pasos de solución

tan^{2} (a) = \frac{( 2 tan ( a ) )}{( 1 - tan ^{2} ( a ) )}

Usando el método de sustitución

tan^{2} (a) = \frac{2 tan ( a )}{1 - tan ^{2} ( a )}

Sea:

tan (a) = u

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

u^{2} = \frac{2 u}{1 - u ^{2}} : u = 0, u \approx - 1.52137 \dots

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

Multiplicar ambos lados por

1 - u^{2}

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

Multiplicar ambos lados por

1 - u^{2}

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

Simplificar

u^{2} (1 - u^{2}) = 2 u

u^{2} (1 - u^{2}) = 2 u

Resolver

u^{2} (1 - u^{2}) = 2 u : u = 0, u \approx - 1.52137 \dots

u^{2} (1 - u^{2}) = 2 u

Mover

2 u

al lado izquierdo

u^{2} (1 - u^{2}) = 2 u

Restar

2 u

de ambos lados

u^{2} (1 - u^{2}) - 2 u = 2 u - 2 u

Simplificar

u^{2} (1 - u^{2}) - 2 u = 0

u^{2} (1 - u^{2}) - 2 u = 0

Factorizar

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

u^{2} (1 - u^{2}) - 2 u

Aplicar las leyes de los exponentes:

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

u^{2} = uu

= uu (- uu + 1) - 2 u

Factorizar el termino común

u

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

Factorizar

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

u (- u^{2} + 1) - 2

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

u (- u^{2} + 1)

Factorizar

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

- u^{2} + 1

Factorizar el termino común

- 1

= - (u^{2} - 1)

Factorizar

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

u^{2} - 1

Reescribir

1

como

1^{2}

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

Aplicar la siguiente regla para binomios al cuadrado:

x^{2} - y^{2} = (x + y) (x - y)

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

= (u + 1) (u - 1)

= - (u + 1) (u - 1)

= - u (u + 1) (u - 1)

= - u (u + 1) (u - 1) - 2

Expandir

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

- u (u + 1) (u - 1) - 2

Expandir

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

Expandir

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

(u + 1) (u - 1)

Aplicar la siguiente regla para binomios al cuadrado:

(a + b) (a - b) = a^{2} - b^{2}

a = u, b = 1

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

Aplicar la regla

1^{a} = 1

1^{2} = 1

= u^{2} - 1

= - u (u^{2} - 1)

Expandir

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

- u (u^{2} - 1)

Poner los parentesis utilizando:

a (b - c) = ab - a c

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

= - u u^{2} - (- u) \cdot 1

Aplicar las reglas de los signos

- (- a) = a

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

Simplificar

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

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

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

u^{2} u

Aplicar las leyes de los exponentes:

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

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

= u^{2 + 1}

Sumar:

2 + 1 = 3

= u^{3}

1 \cdot u = u

1 \cdot u

Multiplicar:

1 \cdot u = u

= u

= - u^{3} + u

= - u^{3} + u

= - u^{3} + u

= - u^{3} + u - 2

= - u^{3} + u - 2

Factorizar

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

- u^{3} + u - 2

Factorizar el termino común

- 1

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

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

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

Simplificar

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

- u (u^{3} - u + 2) = 0

Utilizando el teorema de factor cero: si

ab = 0

entonces

a = 0

b = 0

u = 0 or u^{3} - u + 2 = 0

Resolver

u^{3} - u + 2 = 0 : u \approx - 1.52137 \dots

u^{3} - u + 2 = 0

Encontrar una solución para

u^{3} - u + 2 = 0

utilizando el método de Newton-Raphson:

u \approx - 1.52137 \dots

u^{3} - u + 2 = 0

Definición del método de Newton-Raphson

f (u) = u^{3} - u + 2

Hallar

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

\frac{d}{d u} (u^{3} - u + 2)

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} (2)

\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} (2) = 0

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

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} = - 2 : Δ u_{1} = 1

f (u_{0}) = (- 1)^{3} - (- 1) + 2 = 2

f^{^{'}} (u_{0}) = 3 (- 1)^{2} - 1 = 2

u_{1} = - 2

Δ u_{1} = ∣ - 2 - (- 1) ∣ = 1

Δ u_{1} = 1

u_{2} = - 1.63636 \dots : Δ u_{2} = 0.36363 \dots

f (u_{1}) = (- 2)^{3} - (- 2) + 2 = - 4

f^{^{'}} (u_{1}) = 3 (- 2)^{2} - 1 = 11

u_{2} = - 1.63636 \dots

Δ u_{2} = ∣ - 1.63636 \dots - (- 2) ∣ = 0.36363 \dots

Δ u_{2} = 0.36363 \dots

u_{3} = - 1.53039 \dots : Δ u_{3} = 0.10597 \dots

f (u_{2}) = (- 1.63636 \dots)^{3} - (- 1.63636 \dots) + 2 = - 0.74530 \dots

f^{^{'}} (u_{2}) = 3 (- 1.63636 \dots)^{2} - 1 = 7.03305 \dots

u_{3} = - 1.53039 \dots

Δ u_{3} = ∣ - 1.53039 \dots - (- 1.63636 \dots) ∣ = 0.10597 \dots

Δ u_{3} = 0.10597 \dots

u_{4} = - 1.52144 \dots : Δ u_{4} = 0.00895 \dots

f (u_{3}) = (- 1.53039 \dots)^{3} - (- 1.53039 \dots) + 2 = - 0.05393 \dots

f^{^{'}} (u_{3}) = 3 (- 1.53039 \dots)^{2} - 1 = 6.02629 \dots

u_{4} = - 1.52144 \dots

Δ u_{4} = ∣ - 1.52144 \dots - (- 1.53039 \dots) ∣ = 0.00895 \dots

Δ u_{4} = 0.00895 \dots

u_{5} = - 1.52137 \dots : Δ u_{5} = 0.00006 \dots

f (u_{4}) = (- 1.52144 \dots)^{3} - (- 1.52144 \dots) + 2 = - 0.00036 \dots

f^{^{'}} (u_{4}) = 3 (- 1.52144 \dots)^{2} - 1 = 5.94435 \dots

u_{5} = - 1.52137 \dots

Δ u_{5} = ∣ - 1.52137 \dots - (- 1.52144 \dots) ∣ = 0.00006 \dots

Δ u_{5} = 0.00006 \dots

u_{6} = - 1.52137 \dots : Δ u_{6} = 2.92858 E - 9

f (u_{5}) = (- 1.52137 \dots)^{3} - (- 1.52137 \dots) + 2 = - 1.74069 E - 8

f^{^{'}} (u_{5}) = 3 (- 1.52137 \dots)^{2} - 1 = 5.94378 \dots

u_{6} = - 1.52137 \dots

Δ u_{6} = ∣ - 1.52137 \dots - (- 1.52137 \dots) ∣ = 2.92858 E - 9

Δ u_{6} = 2.92858 E - 9

u \approx - 1.52137 \dots

Aplicar la división larga

Eq u a t i o n 0 : \frac{u ^{3} - u + 2}{u + 1.52137 \dots} = u^{2} - 1.52137 \dots u + 1.31459 \dots

u^{2} - 1.52137 \dots u + 1.31459 \dots \approx 0

Encontrar una solución para

u^{2} - 1.52137 \dots u + 1.31459 \dots = 0

utilizando el método de Newton-Raphson:

Sin solución para

u \in R

u^{2} - 1.52137 \dots u + 1.31459 \dots = 0

Definición del método de Newton-Raphson

f (u) = u^{2} - 1.52137 \dots u + 1.31459 \dots

Hallar

f^{^{'}} (u) : 2 u - 1.52137 \dots

\frac{d}{d u} (u^{2} - 1.52137 \dots u + 1.31459 \dots)

Aplicar la regla de la suma/diferencia:

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

= \frac{d}{d u} (u^{2}) - \frac{d}{d u} (1.52137 \dots u) + \frac{d}{d u} (1.31459 \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} (1.52137 \dots u) = 1.52137 \dots

\frac{d}{d u} (1.52137 \dots u)

Sacar la constante:

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

= 1.52137 \dots \frac{d u}{d u}

Aplicar la regla de derivación:

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

= 1.52137 \dots \cdot 1

Simplificar

= 1.52137 \dots

\frac{d}{d u} (1.31459 \dots) = 0

\frac{d}{d u} (1.31459 \dots)

Derivada de una constante:

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

= 0

= 2 u - 1.52137 \dots + 0

Simplificar

= 2 u - 1.52137 \dots

Sea

u_{0} = 1

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = - 0.65729 \dots : Δ u_{1} = 1.65729 \dots

f (u_{0}) = 1^{2} - 1.52137 \dots \cdot 1 + 1.31459 \dots = 0.79321 \dots

f^{^{'}} (u_{0}) = 2 \cdot 1 - 1.52137 \dots = 0.47862 \dots

u_{1} = - 0.65729 \dots

Δ u_{1} = ∣ - 0.65729 \dots - 1 ∣ = 1.65729 \dots

Δ u_{1} = 1.65729 \dots

u_{2} = 0.31119 \dots : Δ u_{2} = 0.96849 \dots

f (u_{1}) = (- 0.65729 \dots)^{2} - 1.52137 \dots (- 0.65729 \dots) + 1.31459 \dots = 2.74663 \dots

f^{^{'}} (u_{1}) = 2 (- 0.65729 \dots) - 1.52137 \dots = - 2.83597 \dots

u_{2} = 0.31119 \dots

Δ u_{2} = ∣ 0.31119 \dots - (- 0.65729 \dots) ∣ = 0.96849 \dots

Δ u_{2} = 0.96849 \dots

u_{3} = 1.35459 \dots : Δ u_{3} = 1.04339 \dots

f (u_{2}) = 0.31119 \dots^{2} - 1.52137 \dots \cdot 0.31119 \dots + 1.31459 \dots = 0.93798 \dots

f^{^{'}} (u_{2}) = 2 \cdot 0.31119 \dots - 1.52137 \dots = - 0.89897 \dots

u_{3} = 1.35459 \dots

Δ u_{3} = ∣ 1.35459 \dots - 0.31119 \dots ∣ = 1.04339 \dots

Δ u_{3} = 1.04339 \dots

u_{4} = 0.43805 \dots : Δ u_{4} = 0.91653 \dots

f (u_{3}) = 1.35459 \dots^{2} - 1.52137 \dots \cdot 1.35459 \dots + 1.31459 \dots = 1.08866 \dots

f^{^{'}} (u_{3}) = 2 \cdot 1.35459 \dots - 1.52137 \dots = 1.18780 \dots

u_{4} = 0.43805 \dots

Δ u_{4} = ∣ 0.43805 \dots - 1.35459 \dots ∣ = 0.91653 \dots

Δ u_{4} = 0.91653 \dots

u_{5} = 1.73989 \dots : Δ u_{5} = 1.30184 \dots

f (u_{4}) = 0.43805 \dots^{2} - 1.52137 \dots \cdot 0.43805 \dots + 1.31459 \dots = 0.84004 \dots

f^{^{'}} (u_{4}) = 2 \cdot 0.43805 \dots - 1.52137 \dots = - 0.64527 \dots

u_{5} = 1.73989 \dots

Δ u_{5} = ∣ 1.73989 \dots - 0.43805 \dots ∣ = 1.30184 \dots

Δ u_{5} = 1.30184 \dots

u_{6} = 0.87450 \dots : Δ u_{6} = 0.86539 \dots

f (u_{5}) = 1.73989 \dots^{2} - 1.52137 \dots \cdot 1.73989 \dots + 1.31459 \dots = 1.69479 \dots

f^{^{'}} (u_{5}) = 2 \cdot 1.73989 \dots - 1.52137 \dots = 1.95841 \dots

u_{6} = 0.87450 \dots

Δ u_{6} = ∣ 0.87450 \dots - 1.73989 \dots ∣ = 0.86539 \dots

Δ u_{6} = 0.86539 \dots

u_{7} = - 2.41546 \dots : Δ u_{7} = 3.28996 \dots

f (u_{6}) = 0.87450 \dots^{2} - 1.52137 \dots \cdot 0.87450 \dots + 1.31459 \dots = 0.74890 \dots

f^{^{'}} (u_{6}) = 2 \cdot 0.87450 \dots - 1.52137 \dots = 0.22763 \dots

u_{7} = - 2.41546 \dots

Δ u_{7} = ∣ - 2.41546 \dots - 0.87450 \dots ∣ = 3.28996 \dots

Δ u_{7} = 3.28996 \dots

u_{8} = - 0.71153 \dots : Δ u_{8} = 1.70393 \dots

f (u_{7}) = (- 2.41546 \dots)^{2} - 1.52137 \dots (- 2.41546 \dots) + 1.31459 \dots = 10.82387 \dots

f^{^{'}} (u_{7}) = 2 (- 2.41546 \dots) - 1.52137 \dots = - 6.35230 \dots

u_{8} = - 0.71153 \dots

Δ u_{8} = ∣ - 0.71153 \dots - (- 2.41546 \dots) ∣ = 1.70393 \dots

Δ u_{8} = 1.70393 \dots

u_{9} = 0.27452 \dots : Δ u_{9} = 0.98605 \dots

f (u_{8}) = (- 0.71153 \dots)^{2} - 1.52137 \dots (- 0.71153 \dots) + 1.31459 \dots = 2.90337 \dots

f^{^{'}} (u_{8}) = 2 (- 0.71153 \dots) - 1.52137 \dots = - 2.94443 \dots

u_{9} = 0.27452 \dots

Δ u_{9} = ∣ 0.27452 \dots - (- 0.71153 \dots) ∣ = 0.98605 \dots

Δ u_{9} = 0.98605 \dots

u_{10} = 1.27449 \dots : Δ u_{10} = 0.99997 \dots

f (u_{9}) = 0.27452 \dots^{2} - 1.52137 \dots \cdot 0.27452 \dots + 1.31459 \dots = 0.97230 \dots

f^{^{'}} (u_{9}) = 2 \cdot 0.27452 \dots - 1.52137 \dots = - 0.97233 \dots

u_{10} = 1.27449 \dots

Δ u_{10} = ∣ 1.27449 \dots - 0.27452 \dots ∣ = 0.99997 \dots

Δ u_{10} = 0.99997 \dots

No se puede encontrar solución

La solución es

u \approx - 1.52137 \dots

Las soluciones son

u = 0, u \approx - 1.52137 \dots

u = 0, u \approx - 1.52137 \dots

Verificar las soluciones

Encontrar los puntos no definidos (singularidades):

u = 1, u = - 1

Tomar el(los) denominador(es) de

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

y comparar con cero

Resolver

1 - u^{2} = 0 : u = 1, u = - 1

1 - u^{2} = 0

Mover

1

al lado derecho

1 - u^{2} = 0

Restar

1

de ambos lados

1 - u^{2} - 1 = 0 - 1

Simplificar

- u^{2} = - 1

- u^{2} = - 1

Dividir ambos lados entre

- 1

- u^{2} = - 1

Dividir ambos lados entre

- 1

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

Simplificar

u^{2} = 1

u^{2} = 1

Para

x^{2} = f (a)

las soluciones son

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Aplicar las leyes de los exponentes:

1 = 1

= 1

- 1 = - 1

- 1

Aplicar las leyes de los exponentes:

1 = 1

1 = 1

= - 1

u = 1, u = - 1

Los siguientes puntos no están definidos

u = 1, u = - 1

Combinar los puntos no definidos con las soluciones:

u = 0, u \approx - 1.52137 \dots

Sustituir en la ecuación

u = tan (a)

tan (a) = 0, tan (a) \approx - 1.52137 \dots

tan (a) = 0, tan (a) \approx - 1.52137 \dots

tan (a) = 0 : a = πn

tan (a) = 0

Soluciones generales para

tan (a) = 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}

a = 0 + πn

a = 0 + πn

Resolver

a = 0 + πn : a = πn

a = 0 + πn

0 + πn = πn

a = πn

a = πn

tan (a) = - 1.52137 \dots : a = arctan (- 1.52137 \dots) + πn

tan (a) = - 1.52137 \dots

Aplicar propiedades trigonométricas inversas

tan (a) = - 1.52137 \dots

Soluciones generales para

tan (a) = - 1.52137 \dots

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

a = arctan (- 1.52137 \dots) + πn

a = arctan (- 1.52137 \dots) + πn

Combinar toda las soluciones

a = πn, a = arctan (- 1.52137 \dots) + πn

Mostrar soluciones en forma decimal

a = πn, a = - 0.98930 \dots + πn

Ejemplos populares

12cos^2(x)-6=sin(x)

12 cos^{2} (x) - 6 = sin (x)

cos^2(a)= 2/3

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

sin(2x)=-0.848055484

sin (2 x) = - 0.848055484

2+cos^2(x)=3cos(x)

2 + cos^{2} (x) = 3 cos (x)

cos(x+40)=0.85

cos (x + 4 0^{\circ}) = 0.85