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2cos^4(θ)-3cos^2(θ)+1=0

Solución

2 cos^{4} (θ) - 3 cos^{2} (θ) + 1 = 0

Solución

θ = 2 πn, θ = π + 2 πn, θ = 0.78539 \dots + 2 πn, θ = 2 π - 0.78539 \dots + 2 πn, θ = 2.35619 \dots + 2 πn, θ = - 2.35619 \dots + 2 πn

Grados

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n, θ = 4 5^{\circ} + 36 0^{\circ} n, θ = 31 5^{\circ} + 36 0^{\circ} n, θ = 13 5^{\circ} + 36 0^{\circ} n, θ = - 13 5^{\circ} + 36 0^{\circ} n

Pasos de solución

2 cos^{4} (θ) - 3 cos^{2} (θ) + 1 = 0

Usando el método de sustitución

2 cos^{4} (θ) - 3 cos^{2} (θ) + 1 = 0

Sea:

cos (θ) = u

2 u^{4} - 3 u^{2} + 1 = 0

2 u^{4} - 3 u^{2} + 1 = 0 : u = 1, u = - 1, u = \frac{1}{2}, u = - \frac{1}{2}

2 u^{4} - 3 u^{2} + 1 = 0

Re-escribir la ecuación con

v = u^{2}

v^{2} = u^{4}

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

Resolver

2 v^{2} - 3 v + 1 = 0 : v = 1, v = \frac{1}{2}

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

Resolver con la fórmula general para ecuaciones de segundo grado:

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

Formula general para ecuaciones de segundo grado:

Para

a = 2, b = - 3, c = 1

v_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

v_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

(- 3)^{2} - 4 \cdot 2 \cdot 1 = 1

(- 3)^{2} - 4 \cdot 2 \cdot 1

Aplicar las leyes de los exponentes:

(- a)^{n} = a^{n},

n

es par

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

= 3^{2} - 4 \cdot 2 \cdot 1

Multiplicar los numeros:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Restar:

9 - 8 = 1

= 1

Aplicar la regla

1 = 1

= 1

v_{1, 2} = \frac{- ( - 3 ) \pm 1}{2 \cdot 2}

Separar las soluciones

v_{1} = \frac{- ( - 3 ) + 1}{2 \cdot 2}, v_{2} = \frac{- ( - 3 ) - 1}{2 \cdot 2}

v = \frac{- ( - 3 ) + 1}{2 \cdot 2} : 1

\frac{- ( - 3 ) + 1}{2 \cdot 2}

Aplicar la regla

- (- a) = a

= \frac{3 + 1}{2 \cdot 2}

Sumar:

3 + 1 = 4

= \frac{4}{2 \cdot 2}

Multiplicar los numeros:

2 \cdot 2 = 4

= \frac{4}{4}

Aplicar la regla

\frac{a}{a} = 1

= 1

v = \frac{- ( - 3 ) - 1}{2 \cdot 2} : \frac{1}{2}

\frac{- ( - 3 ) - 1}{2 \cdot 2}

Aplicar la regla

- (- a) = a

= \frac{3 - 1}{2 \cdot 2}

Restar:

3 - 1 = 2

= \frac{2}{2 \cdot 2}

Multiplicar los numeros:

2 \cdot 2 = 4

= \frac{2}{4}

Eliminar los terminos comunes:

2

= \frac{1}{2}

Las soluciones a la ecuación de segundo grado son:

v = 1, v = \frac{1}{2}

v = 1, v = \frac{1}{2}

Sustituir hacia atrás la

v = u^{2},

resolver para

u

Resolver

u^{2} = 1 : u = 1, u = - 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 la regla

1 = 1

= 1

- 1 = - 1

- 1

Aplicar la regla

1 = 1

= - 1

u = 1, u = - 1

Resolver

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

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

Para

x^{2} = f (a)

las soluciones son

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

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

Las soluciones son

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

Sustituir en la ecuación

u = cos (θ)

cos (θ) = 1, cos (θ) = - 1, cos (θ) = \frac{1}{2}, cos (θ) = - \frac{1}{2}

cos (θ) = 1, cos (θ) = - 1, cos (θ) = \frac{1}{2}, cos (θ) = - \frac{1}{2}

cos (θ) = 1 : θ = 2 πn

cos (θ) = 1

Soluciones generales para

cos (θ) = 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}

θ = 0 + 2 πn

θ = 0 + 2 πn

Resolver

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

cos (θ) = - 1 : θ = π + 2 πn

cos (θ) = - 1

Soluciones generales para

cos (θ) = - 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}

θ = π + 2 πn

θ = π + 2 πn

cos (θ) = \frac{1}{2} : θ = arccos (\frac{1}{2}) + 2 πn, θ = 2 π - arccos (\frac{1}{2}) + 2 πn

cos (θ) = \frac{1}{2}

Aplicar propiedades trigonométricas inversas

cos (θ) = \frac{1}{2}

Soluciones generales para

cos (θ) = \frac{1}{2}

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

θ = arccos (\frac{1}{2}) + 2 πn, θ = 2 π - arccos (\frac{1}{2}) + 2 πn

θ = arccos (\frac{1}{2}) + 2 πn, θ = 2 π - arccos (\frac{1}{2}) + 2 πn

cos (θ) = - \frac{1}{2} : θ = arccos (- \frac{1}{2}) + 2 πn, θ = - arccos (- \frac{1}{2}) + 2 πn

cos (θ) = - \frac{1}{2}

Aplicar propiedades trigonométricas inversas

cos (θ) = - \frac{1}{2}

Soluciones generales para

cos (θ) = - \frac{1}{2}

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

θ = arccos (- \frac{1}{2}) + 2 πn, θ = - arccos (- \frac{1}{2}) + 2 πn

θ = arccos (- \frac{1}{2}) + 2 πn, θ = - arccos (- \frac{1}{2}) + 2 πn

Combinar toda las soluciones

θ = 2 πn, θ = π + 2 πn, θ = arccos (\frac{1}{2}) + 2 πn, θ = 2 π - arccos (\frac{1}{2}) + 2 πn, θ = arccos (- \frac{1}{2}) + 2 πn, θ = - arccos (- \frac{1}{2}) + 2 πn

Mostrar soluciones en forma decimal

θ = 2 πn, θ = π + 2 πn, θ = 0.78539 \dots + 2 πn, θ = 2 π - 0.78539 \dots + 2 πn, θ = 2.35619 \dots + 2 πn, θ = - 2.35619 \dots + 2 πn

Ejemplos populares

2cos^2(θ)+7cos(θ)-4=0

2 cos^{2} (θ) + 7 cos (θ) - 4 = 0

sec^4(x)=sec^2(x)tan^2(x)-2tan^4(x)

sec^{4} (x) = sec^{2} (x) tan^{2} (x) - 2 tan^{4} (x)

sin(x)cos(2x)+cos(x)sin(2x)= 1/(sqrt(2))

sin (x) cos (2 x) + cos (x) sin (2 x) = \frac{1}{2}

sin((5pi)/6-2x)=cos(x-pi/6),sin((2pi)/3-x)

sin (\frac{5 π}{6} - 2 x) = cos (x - \frac{π}{6}), sin (\frac{2 π}{3} - x)

0.08=0.1cos(4x-1.57)

0.08 = 0.1 cos (4 x - 1.57)