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

Solución

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

Solución

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 2 πn, x = 0.99597 \dots + 2 πn, x = 2 π - 0.99597 \dots + 2 πn

Grados

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n, x = 57.06487 \dots^{\circ} + 36 0^{\circ} n, x = 302.93512 \dots^{\circ} + 36 0^{\circ} n

Pasos de solución

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

Usando el método de sustitución

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

Sea:

cos (x) = u

u^{5} + u = 2 u^{2}

u^{5} + u = 2 u^{2} : u = 0, u = 1, u \approx 0.54368 \dots

u^{5} + u = 2 u^{2}

Mover

2 u^{2}

al lado izquierdo

u^{5} + u = 2 u^{2}

Restar

2 u^{2}

de ambos lados

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

Simplificar

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

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

Factorizar

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

u^{5} + u - 2 u^{2}

Factorizar el termino común

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

u^{5} - 2 u^{2} + u

Aplicar las leyes de los exponentes:

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

u^{2} = uu

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

Factorizar el termino común

u

= u (u^{4} - 2 u + 1)

= u (u^{4} - 2 u + 1)

Factorizar

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

u^{4} - 2 u + 1

Utilizar el teorema de la raíz racional

a_{0} = 1, a_{n} = 1

Los divisores de

a_{0} : 1,

Los divisores de

a_{n} : 1

Por lo tanto, verificar los siguientes numeros racionales:

\pm \frac{1}{1}

\frac{1}{1}

es la raíz de la expresión, por lo tanto, factorizar

u - 1

= (u - 1) \frac{u ^{4} - 2 u + 1}{u - 1}

\frac{u ^{4} - 2 u + 1}{u - 1} = u^{3} + u^{2} + u - 1

\frac{u ^{4} - 2 u + 1}{u - 1}

Dividir

\frac{u ^{4} - 2 u + 1}{u - 1} : \frac{u ^{4} - 2 u + 1}{u - 1} = u^{3} + \frac{u ^{3} - 2 u + 1}{u - 1}

Dividir los coeficientes de los términos de mayor grado del numerador

u^{4} - 2 u + 1

y el divisor

u - 1 : \frac{u ^{4}}{u} = u^{3}

Cociente = u^{3}

Multiplicar

u - 1

por

u^{3} : u^{4} - u^{3}

Substraer

u^{4} - u^{3}

u^{4} - 2 u + 1

para obtener un nuevo residuo

Residuo = u^{3} - 2 u + 1

Por lo tanto

\frac{u ^{4} - 2 u + 1}{u - 1} = u^{3} + \frac{u ^{3} - 2 u + 1}{u - 1}

= u^{3} + \frac{u ^{3} - 2 u + 1}{u - 1}

Dividir

\frac{u ^{3} - 2 u + 1}{u - 1} : \frac{u ^{3} - 2 u + 1}{u - 1} = u^{2} + \frac{u ^{2} - 2 u + 1}{u - 1}

Dividir los coeficientes de los términos de mayor grado del numerador

u^{3} - 2 u + 1

y el divisor

u - 1 : \frac{u ^{3}}{u} = u^{2}

Cociente = u^{2}

Multiplicar

u - 1

por

u^{2} : u^{3} - u^{2}

Substraer

u^{3} - u^{2}

u^{3} - 2 u + 1

para obtener un nuevo residuo

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

Por lo tanto

\frac{u ^{3} - 2 u + 1}{u - 1} = u^{2} + \frac{u ^{2} - 2 u + 1}{u - 1}

= u^{3} + u^{2} + \frac{u ^{2} - 2 u + 1}{u - 1}

Dividir

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

Dividir los coeficientes de los términos de mayor grado del numerador

u^{2} - 2 u + 1

y el divisor

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

Cociente = u

Multiplicar

u - 1

por

u : u^{2} - u

Substraer

u^{2} - u

u^{2} - 2 u + 1

para obtener un nuevo residuo

Residuo = - u + 1

Por lo tanto

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

= u^{3} + u^{2} + u + \frac{- u + 1}{u - 1}

Dividir

\frac{- u + 1}{u - 1} : \frac{- u + 1}{u - 1} = - 1

Dividir los coeficientes de los términos de mayor grado del numerador

- u + 1

y el divisor

u - 1 : \frac{- u}{u} = - 1

Cociente = - 1

Multiplicar

u - 1

por

- 1 : - u + 1

Substraer

- u + 1

- u + 1

para obtener un nuevo residuo

Residuo = 0

Por lo tanto

\frac{- u + 1}{u - 1} = - 1

= u^{3} + u^{2} + u - 1

= u^{3} + u^{2} + u - 1

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

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

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

Utilizando el teorema de factor cero: si

ab = 0

entonces

a = 0

b = 0

u = 0 or u - 1 = 0 or u^{3} + u^{2} + u - 1 = 0

Resolver

u - 1 = 0 : u = 1

u - 1 = 0

Mover

1

al lado derecho

u - 1 = 0

Sumar

1

a ambos lados

u - 1 + 1 = 0 + 1

Simplificar

u = 1

u = 1

Resolver

u^{3} + u^{2} + u - 1 = 0 : u \approx 0.54368 \dots

u^{3} + u^{2} + u - 1 = 0

Encontrar una solución para

u^{3} + u^{2} + u - 1 = 0

utilizando el método de Newton-Raphson:

u \approx 0.54368 \dots

u^{3} + u^{2} + u - 1 = 0

Definición del método de Newton-Raphson

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

Hallar

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

\frac{d}{d u} (u^{3} + u^{2} + u - 1)

Aplicar la regla de la suma/diferencia:

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

= \frac{d}{d u} (u^{3}) + \frac{d}{d u} (u^{2}) + \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}{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 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} + 2 u + 1 - 0

Simplificar

= 3 u^{2} + 2 u + 1

Sea

u_{0} = 1

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = 0.66666 \dots : Δ u_{1} = 0.33333 \dots

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

f^{^{'}} (u_{0}) = 3 \cdot 1^{2} + 2 \cdot 1 + 1 = 6

u_{1} = 0.66666 \dots

Δ u_{1} = ∣ 0.66666 \dots - 1 ∣ = 0.33333 \dots

Δ u_{1} = 0.33333 \dots

u_{2} = 0.55555 \dots : Δ u_{2} = 0.11111 \dots

f (u_{1}) = 0.66666 \dots^{3} + 0.66666 \dots^{2} + 0.66666 \dots - 1 = 0.40740 \dots

f^{^{'}} (u_{1}) = 3 \cdot 0.66666 \dots^{2} + 2 \cdot 0.66666 \dots + 1 = 3.66666 \dots

u_{2} = 0.55555 \dots

Δ u_{2} = ∣ 0.55555 \dots - 0.66666 \dots ∣ = 0.11111 \dots

Δ u_{2} = 0.11111 \dots

u_{3} = 0.54381 \dots : Δ u_{3} = 0.01174 \dots

f (u_{2}) = 0.55555 \dots^{3} + 0.55555 \dots^{2} + 0.55555 \dots - 1 = 0.03566 \dots

f^{^{'}} (u_{2}) = 3 \cdot 0.55555 \dots^{2} + 2 \cdot 0.55555 \dots + 1 = 3.03703 \dots

u_{3} = 0.54381 \dots

Δ u_{3} = ∣ 0.54381 \dots - 0.55555 \dots ∣ = 0.01174 \dots

Δ u_{3} = 0.01174 \dots

u_{4} = 0.54368 \dots : Δ u_{4} = 0.00012 \dots

f (u_{3}) = 0.54381 \dots^{3} + 0.54381 \dots^{2} + 0.54381 \dots - 1 = 0.00036 \dots

f^{^{'}} (u_{3}) = 3 \cdot 0.54381 \dots^{2} + 2 \cdot 0.54381 \dots + 1 = 2.97481 \dots

u_{4} = 0.54368 \dots

Δ u_{4} = ∣ 0.54368 \dots - 0.54381 \dots ∣ = 0.00012 \dots

Δ u_{4} = 0.00012 \dots

u_{5} = 0.54368 \dots : Δ u_{5} = 1.34021 E - 8

f (u_{4}) = 0.54368 \dots^{3} + 0.54368 \dots^{2} + 0.54368 \dots - 1 = 3.98601 E - 8

f^{^{'}} (u_{4}) = 3 \cdot 0.54368 \dots^{2} + 2 \cdot 0.54368 \dots + 1 = 2.97417 \dots

u_{5} = 0.54368 \dots

Δ u_{5} = ∣ 0.54368 \dots - 0.54368 \dots ∣ = 1.34021 E - 8

Δ u_{5} = 1.34021 E - 8

u \approx 0.54368 \dots

Aplicar la división larga

Eq u a t i o n 0 : \frac{u ^{3} + u ^{2} + u - 1}{u - 0.54368 \dots} = u^{2} + 1.54368 \dots u + 1.83928 \dots

u^{2} + 1.54368 \dots u + 1.83928 \dots \approx 0

Encontrar una solución para

u^{2} + 1.54368 \dots u + 1.83928 \dots = 0

utilizando el método de Newton-Raphson:

Sin solución para

u \in R

u^{2} + 1.54368 \dots u + 1.83928 \dots = 0

Definición del método de Newton-Raphson

f (u) = u^{2} + 1.54368 \dots u + 1.83928 \dots

Hallar

f^{^{'}} (u) : 2 u + 1.54368 \dots

\frac{d}{d u} (u^{2} + 1.54368 \dots u + 1.83928 \dots)

Aplicar la regla de la suma/diferencia:

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

= \frac{d}{d u} (u^{2}) + \frac{d}{d u} (1.54368 \dots u) + \frac{d}{d u} (1.83928 \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} (1.54368 \dots u) = 1.54368 \dots

\frac{d}{d u} (1.54368 \dots u)

Sacar la constante:

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

= 1.54368 \dots \frac{d u}{d u}

Aplicar la regla de derivación:

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

= 1.54368 \dots \cdot 1

Simplificar

= 1.54368 \dots

\frac{d}{d u} (1.83928 \dots) = 0

\frac{d}{d u} (1.83928 \dots)

Derivada de una constante:

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

= 0

= 2 u + 1.54368 \dots + 0

Simplificar

= 2 u + 1.54368 \dots

Sea

u_{0} = - 1

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = 1.83928 \dots : Δ u_{1} = 2.83928 \dots

f (u_{0}) = (- 1)^{2} + 1.54368 \dots (- 1) + 1.83928 \dots = 1.29559 \dots

f^{^{'}} (u_{0}) = 2 (- 1) + 1.54368 \dots = - 0.45631 \dots

u_{1} = 1.83928 \dots

Δ u_{1} = ∣ 1.83928 \dots - (- 1) ∣ = 2.83928 \dots

Δ u_{1} = 2.83928 \dots

u_{2} = 0.29559 \dots : Δ u_{2} = 1.54368 \dots

f (u_{1}) = 1.83928 \dots^{2} + 1.54368 \dots \cdot 1.83928 \dots + 1.83928 \dots = 8.06154 \dots

f^{^{'}} (u_{1}) = 2 \cdot 1.83928 \dots + 1.54368 \dots = 5.22226 \dots

u_{2} = 0.29559 \dots

Δ u_{2} = ∣ 0.29559 \dots - 1.83928 \dots ∣ = 1.54368 \dots

Δ u_{2} = 1.54368 \dots

u_{3} = - 0.82061 \dots : Δ u_{3} = 1.11620 \dots

f (u_{2}) = 0.29559 \dots^{2} + 1.54368 \dots \cdot 0.29559 \dots + 1.83928 \dots = 2.38297 \dots

f^{^{'}} (u_{2}) = 2 \cdot 0.29559 \dots + 1.54368 \dots = 2.13488 \dots

u_{3} = - 0.82061 \dots

Δ u_{3} = ∣ - 0.82061 \dots - 0.29559 \dots ∣ = 1.11620 \dots

Δ u_{3} = 1.11620 \dots

u_{4} = 11.95386 \dots : Δ u_{4} = 12.77447 \dots

f (u_{3}) = (- 0.82061 \dots)^{2} + 1.54368 \dots (- 0.82061 \dots) + 1.83928 \dots = 1.24592 \dots

f^{^{'}} (u_{3}) = 2 (- 0.82061 \dots) + 1.54368 \dots = - 0.09753 \dots

u_{4} = 11.95386 \dots

Δ u_{4} = ∣ 11.95386 \dots - (- 0.82061 \dots) ∣ = 12.77447 \dots

Δ u_{4} = 12.77447 \dots

u_{5} = 5.54214 \dots : Δ u_{5} = 6.41171 \dots

f (u_{4}) = 11.95386 \dots^{2} + 1.54368 \dots \cdot 11.95386 \dots + 1.83928 \dots = 163.18717 \dots

f^{^{'}} (u_{4}) = 2 \cdot 11.95386 \dots + 1.54368 \dots = 25.45141 \dots

u_{5} = 5.54214 \dots

Δ u_{5} = ∣ 5.54214 \dots - 11.95386 \dots ∣ = 6.41171 \dots

Δ u_{5} = 6.41171 \dots

u_{6} = 2.28667 \dots : Δ u_{6} = 3.25547 \dots

f (u_{5}) = 5.54214 \dots^{2} + 1.54368 \dots \cdot 5.54214 \dots + 1.83928 \dots = 41.11006 \dots

f^{^{'}} (u_{5}) = 2 \cdot 5.54214 \dots + 1.54368 \dots = 12.62798 \dots

u_{6} = 2.28667 \dots

Δ u_{6} = ∣ 2.28667 \dots - 5.54214 \dots ∣ = 3.25547 \dots

Δ u_{6} = 3.25547 \dots

u_{7} = 0.55412 \dots : Δ u_{7} = 1.73255 \dots

f (u_{6}) = 2.28667 \dots^{2} + 1.54368 \dots \cdot 2.28667 \dots + 1.83928 \dots = 10.59809 \dots

f^{^{'}} (u_{6}) = 2 \cdot 2.28667 \dots + 1.54368 \dots = 6.11704 \dots

u_{7} = 0.55412 \dots

Δ u_{7} = ∣ 0.55412 \dots - 2.28667 \dots ∣ = 1.73255 \dots

Δ u_{7} = 1.73255 \dots

u_{8} = - 0.57777 \dots : Δ u_{8} = 1.13190 \dots

f (u_{7}) = 0.55412 \dots^{2} + 1.54368 \dots \cdot 0.55412 \dots + 1.83928 \dots = 3.00173 \dots

f^{^{'}} (u_{7}) = 2 \cdot 0.55412 \dots + 1.54368 \dots = 2.65193 \dots

u_{8} = - 0.57777 \dots

Δ u_{8} = ∣ - 0.57777 \dots - 0.55412 \dots ∣ = 1.13190 \dots

Δ u_{8} = 1.13190 \dots

u_{9} = - 3.87872 \dots : Δ u_{9} = 3.30094 \dots

f (u_{8}) = (- 0.57777 \dots)^{2} + 1.54368 \dots (- 0.57777 \dots) + 1.83928 \dots = 1.28120 \dots

f^{^{'}} (u_{8}) = 2 (- 0.57777 \dots) + 1.54368 \dots = 0.38813 \dots

u_{9} = - 3.87872 \dots

Δ u_{9} = ∣ - 3.87872 \dots - (- 0.57777 \dots) ∣ = 3.30094 \dots

Δ u_{9} = 3.30094 \dots

u_{10} = - 2.12515 \dots : Δ u_{10} = 1.75356 \dots

f (u_{9}) = (- 3.87872 \dots)^{2} + 1.54368 \dots (- 3.87872 \dots) + 1.83928 \dots = 10.89622 \dots

f^{^{'}} (u_{9}) = 2 (- 3.87872 \dots) + 1.54368 \dots = - 6.21375 \dots

u_{10} = - 2.12515 \dots

Δ u_{10} = ∣ - 2.12515 \dots - (- 3.87872 \dots) ∣ = 1.75356 \dots

Δ u_{10} = 1.75356 \dots

u_{11} = - 0.98905 \dots : Δ u_{11} = 1.13610 \dots

f (u_{10}) = (- 2.12515 \dots)^{2} + 1.54368 \dots (- 2.12515 \dots) + 1.83928 \dots = 3.07499 \dots

f^{^{'}} (u_{10}) = 2 (- 2.12515 \dots) + 1.54368 \dots = - 2.70662 \dots

u_{11} = - 0.98905 \dots

Δ u_{11} = ∣ - 0.98905 \dots - (- 2.12515 \dots) ∣ = 1.13610 \dots

Δ u_{11} = 1.13610 \dots

u_{12} = 1.98207 \dots : Δ u_{12} = 2.97113 \dots

f (u_{11}) = (- 0.98905 \dots)^{2} + 1.54368 \dots (- 0.98905 \dots) + 1.83928 \dots = 1.29072 \dots

f^{^{'}} (u_{11}) = 2 (- 0.98905 \dots) + 1.54368 \dots = - 0.43442 \dots

u_{12} = 1.98207 \dots

Δ u_{12} = ∣ 1.98207 \dots - (- 0.98905 \dots) ∣ = 2.97113 \dots

Δ u_{12} = 2.97113 \dots

u_{13} = 0.37934 \dots : Δ u_{13} = 1.60273 \dots

f (u_{12}) = 1.98207 \dots^{2} + 1.54368 \dots \cdot 1.98207 \dots + 1.83928 \dots = 8.82763 \dots

f^{^{'}} (u_{12}) = 2 \cdot 1.98207 \dots + 1.54368 \dots = 5.50784 \dots

u_{13} = 0.37934 \dots

Δ u_{13} = ∣ 0.37934 \dots - 1.98207 \dots ∣ = 1.60273 \dots

Δ u_{13} = 1.60273 \dots

u_{14} = - 0.73636 \dots : Δ u_{14} = 1.11570 \dots

f (u_{13}) = 0.37934 \dots^{2} + 1.54368 \dots \cdot 0.37934 \dots + 1.83928 \dots = 2.56876 \dots

f^{^{'}} (u_{13}) = 2 \cdot 0.37934 \dots + 1.54368 \dots = 2.30236 \dots

u_{14} = - 0.73636 \dots

Δ u_{14} = ∣ - 0.73636 \dots - 0.37934 \dots ∣ = 1.11570 \dots

Δ u_{14} = 1.11570 \dots

u_{15} = - 18.27956 \dots : Δ u_{15} = 17.54320 \dots

f (u_{14}) = (- 0.73636 \dots)^{2} + 1.54368 \dots (- 0.73636 \dots) + 1.83928 \dots = 1.24480 \dots

f^{^{'}} (u_{14}) = 2 (- 0.73636 \dots) + 1.54368 \dots = 0.07095 \dots

u_{15} = - 18.27956 \dots

Δ u_{15} = ∣ - 18.27956 \dots - (- 0.73636 \dots) ∣ = 17.54320 \dots

Δ u_{15} = 17.54320 \dots

No se puede encontrar solución

La solución es

u \approx 0.54368 \dots

Las soluciones son

u = 0, u = 1, u \approx 0.54368 \dots

Sustituir en la ecuación

u = cos (x)

cos (x) = 0, cos (x) = 1, cos (x) \approx 0.54368 \dots

cos (x) = 0, cos (x) = 1, cos (x) \approx 0.54368 \dots

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

Soluciones generales para

cos (x) = 0

cos (x)

tabla de valores periódicos con

2 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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 1 : x = 2 πn

cos (x) = 1

Soluciones generales para

cos (x) = 1

cos (x)

tabla de valores periódicos con

2 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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = 0 + 2 πn

x = 0 + 2 πn

Resolver

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

cos (x) = 0.54368 \dots : x = arccos (0.54368 \dots) + 2 πn, x = 2 π - arccos (0.54368 \dots) + 2 πn

cos (x) = 0.54368 \dots

Aplicar propiedades trigonométricas inversas

cos (x) = 0.54368 \dots

Soluciones generales para

cos (x) = 0.54368 \dots

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (0.54368 \dots) + 2 πn, x = 2 π - arccos (0.54368 \dots) + 2 πn

x = arccos (0.54368 \dots) + 2 πn, x = 2 π - arccos (0.54368 \dots) + 2 πn

Combinar toda las soluciones

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 2 πn, x = arccos (0.54368 \dots) + 2 πn, x = 2 π - arccos (0.54368 \dots) + 2 πn

Mostrar soluciones en forma decimal

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 2 πn, x = 0.99597 \dots + 2 πn, x = 2 π - 0.99597 \dots + 2 πn

Ejemplos populares

sin(40+x)=cos(5x+10)

sin (4 0^{\circ} + x) = cos (5 x + 10)

cos(a)= 5/8

cos (a) = \frac{5}{8}

tan(θ)sin^2(θ)=tan(θ)

tan (θ) sin^{2} (θ) = tan (θ)

cos(θ)= 5/8

cos (θ) = \frac{5}{8}

6(1-cos(θ))=sin^2(θ),0<= θ<= 2pi

6 (1 - cos (θ)) = sin^{2} (θ), 0 \leq θ \leq 2 π