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1.04cos^2(θ)+0.102cos(θ)-0.935=0

Solución

1.04 cos^{2} (θ) + 0.102 cos (θ) - 0.935 = 0

Solución

θ = 0.45009 \dots + 2 πn, θ = 2 π - 0.45009 \dots + 2 πn, θ = 3.08648 \dots + 2 πn, θ = - 3.08648 \dots + 2 πn

Grados

θ = 25.78862 \dots^{\circ} + 36 0^{\circ} n, θ = 334.21137 \dots^{\circ} + 36 0^{\circ} n, θ = 176.84271 \dots^{\circ} + 36 0^{\circ} n, θ = - 176.84271 \dots^{\circ} + 36 0^{\circ} n

Pasos de solución

1.04 cos^{2} (θ) + 0.102 cos (θ) - 0.935 = 0

Usando el método de sustitución

1.04 cos^{2} (θ) + 0.102 cos (θ) - 0.935 = 0

Sea:

cos (θ) = u

1.04 u^{2} + 0.102 u - 0.935 = 0

1.04 u^{2} + 0.102 u - 0.935 = 0 : u = \frac{- 51 + 975001}{1040}, u = - \frac{51 + 975001}{1040}

1.04 u^{2} + 0.102 u - 0.935 = 0

Multiplicar ambos lados por

1000

1.04 u^{2} + 0.102 u - 0.935 = 0

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere are

3

digits to the right of the decimal point, therefore multiply by

1000

1.04 u^{2} \cdot 1000 + 0.102 u \cdot 1000 - 0.935 \cdot 1000 = 0 \cdot 1000

Simplificar

1040 u^{2} + 102 u - 935 = 0

1040 u^{2} + 102 u - 935 = 0

Resolver con la fórmula general para ecuaciones de segundo grado:

1040 u^{2} + 102 u - 935 = 0

Formula general para ecuaciones de segundo grado:

Para

a = 1040, b = 102, c = - 935

u_{1, 2} = \frac{- 102 \pm 10 2 ^{2} - 4 \cdot 1040 ( - 935 )}{2 \cdot 1040}

u_{1, 2} = \frac{- 102 \pm 10 2 ^{2} - 4 \cdot 1040 ( - 935 )}{2 \cdot 1040}

10 2^{2} - 4 \cdot 1040 (- 935) = 2975001

10 2^{2} - 4 \cdot 1040 (- 935)

Aplicar la regla

- (- a) = a

= 10 2^{2} + 4 \cdot 1040 \cdot 935

Multiplicar los numeros:

4 \cdot 1040 \cdot 935 = 3889600

= 10 2^{2} + 3889600

10 2^{2} = 10404

= 10404 + 3889600

Sumar:

10404 + 3889600 = 3900004

= 3900004

Descomposición en factores primos de

3900004 : 2^{2} \cdot 17 \cdot 83 \cdot 691

3900004

= 2^{2} \cdot 17 \cdot 83 \cdot 691

Aplicar las leyes de los exponentes:

= 2^{2} 17 \cdot 83 \cdot 691

Aplicar las leyes de los exponentes:

2^{2} = 2

= 2 17 \cdot 83 \cdot 691

Simplificar

= 2975001

u_{1, 2} = \frac{- 102 \pm 2 975001}{2 \cdot 1040}

Separar las soluciones

u_{1} = \frac{- 102 + 2 975001}{2 \cdot 1040}, u_{2} = \frac{- 102 - 2 975001}{2 \cdot 1040}

u = \frac{- 102 + 2 975001}{2 \cdot 1040} : \frac{- 51 + 975001}{1040}

\frac{- 102 + 2 975001}{2 \cdot 1040}

Multiplicar los numeros:

2 \cdot 1040 = 2080

= \frac{- 102 + 2 975001}{2080}

Factorizar

- 102 + 2975001 : 2 (- 51 + 975001)

- 102 + 2975001

Reescribir como

= - 2 \cdot 51 + 2975001

Factorizar el termino común

2

= 2 (- 51 + 975001)

= \frac{2 ( - 51 + 975001 )}{2080}

Eliminar los terminos comunes:

2

= \frac{- 51 + 975001}{1040}

u = \frac{- 102 - 2 975001}{2 \cdot 1040} : - \frac{51 + 975001}{1040}

\frac{- 102 - 2 975001}{2 \cdot 1040}

Multiplicar los numeros:

2 \cdot 1040 = 2080

= \frac{- 102 - 2 975001}{2080}

Factorizar

- 102 - 2975001 : - 2 (51 + 975001)

- 102 - 2975001

Reescribir como

= - 2 \cdot 51 - 2975001

Factorizar el termino común

2

= - 2 (51 + 975001)

= - \frac{2 ( 51 + 975001 )}{2080}

Eliminar los terminos comunes:

2

= - \frac{51 + 975001}{1040}

Las soluciones a la ecuación de segundo grado son:

u = \frac{- 51 + 975001}{1040}, u = - \frac{51 + 975001}{1040}

Sustituir en la ecuación

u = cos (θ)

cos (θ) = \frac{- 51 + 975001}{1040}, cos (θ) = - \frac{51 + 975001}{1040}

cos (θ) = \frac{- 51 + 975001}{1040}, cos (θ) = - \frac{51 + 975001}{1040}

cos (θ) = \frac{- 51 + 975001}{1040} : θ = arccos (\frac{- 51 + 975001}{1040}) + 2 πn, θ = 2 π - arccos (\frac{- 51 + 975001}{1040}) + 2 πn

cos (θ) = \frac{- 51 + 975001}{1040}

Aplicar propiedades trigonométricas inversas

cos (θ) = \frac{- 51 + 975001}{1040}

Soluciones generales para

cos (θ) = \frac{- 51 + 975001}{1040}

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

θ = arccos (\frac{- 51 + 975001}{1040}) + 2 πn, θ = 2 π - arccos (\frac{- 51 + 975001}{1040}) + 2 πn

θ = arccos (\frac{- 51 + 975001}{1040}) + 2 πn, θ = 2 π - arccos (\frac{- 51 + 975001}{1040}) + 2 πn

cos (θ) = - \frac{51 + 975001}{1040} : θ = arccos (- \frac{51 + 975001}{1040}) + 2 πn, θ = - arccos (- \frac{51 + 975001}{1040}) + 2 πn

cos (θ) = - \frac{51 + 975001}{1040}

Aplicar propiedades trigonométricas inversas

cos (θ) = - \frac{51 + 975001}{1040}

Soluciones generales para

cos (θ) = - \frac{51 + 975001}{1040}

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

θ = arccos (- \frac{51 + 975001}{1040}) + 2 πn, θ = - arccos (- \frac{51 + 975001}{1040}) + 2 πn

θ = arccos (- \frac{51 + 975001}{1040}) + 2 πn, θ = - arccos (- \frac{51 + 975001}{1040}) + 2 πn

Combinar toda las soluciones

θ = arccos (\frac{- 51 + 975001}{1040}) + 2 πn, θ = 2 π - arccos (\frac{- 51 + 975001}{1040}) + 2 πn, θ = arccos (- \frac{51 + 975001}{1040}) + 2 πn, θ = - arccos (- \frac{51 + 975001}{1040}) + 2 πn

Mostrar soluciones en forma decimal

θ = 0.45009 \dots + 2 πn, θ = 2 π - 0.45009 \dots + 2 πn, θ = 3.08648 \dots + 2 πn, θ = - 3.08648 \dots + 2 πn

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