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(cos^2(a)-3cos(a)+2)/(sin^2(a))=1

Solución

\frac{cos ^{2} ( a ) - 3 cos ( a ) + 2}{sin ^{2} ( a )} = 1

Solución

a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn

Grados

a = 6 0^{\circ} + 36 0^{\circ} n, a = 30 0^{\circ} + 36 0^{\circ} n

Pasos de solución

\frac{cos ^{2} ( a ) - 3 cos ( a ) + 2}{sin ^{2} ( a )} = 1

Restar

1

de ambos lados

\frac{cos ^{2} ( a ) - 3 cos ( a ) + 2}{sin ^{2} ( a )} - 1 = 0

Simplificar

\frac{cos ^{2} ( a ) - 3 cos ( a ) + 2}{sin ^{2} ( a )} - 1 : \frac{cos ^{2} ( a ) - 3 cos ( a ) + 2 - sin ^{2} ( a )}{sin ^{2} ( a )}

\frac{cos ^{2} ( a ) - 3 cos ( a ) + 2}{sin ^{2} ( a )} - 1

Convertir a fracción:

1 = \frac{1 sin ^{2} ( a )}{sin ^{2} ( a )}

= \frac{cos ^{2} ( a ) - 3 cos ( a ) + 2}{sin ^{2} ( a )} - \frac{1 \cdot sin ^{2} ( a )}{sin ^{2} ( a )}

Ya que los denominadores son iguales, combinar las fracciones:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{cos ^{2} ( a ) - 3 cos ( a ) + 2 - 1 \cdot sin ^{2} ( a )}{sin ^{2} ( a )}

Multiplicar:

1 \cdot sin^{2} (a) = sin^{2} (a)

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

\frac{cos ^{2} ( a ) - 3 cos ( a ) + 2 - sin ^{2} ( a )}{sin ^{2} ( a )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

cos^{2} (a) - 3 cos (a) + 2 - sin^{2} (a) = 0

Re-escribir usando identidades trigonométricas

2 + cos^{2} (a) - sin^{2} (a) - 3 cos (a)

Utilizar la identidad pitagórica:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= 2 + cos^{2} (a) - (1 - cos^{2} (a)) - 3 cos (a)

Simplificar

2 + cos^{2} (a) - (1 - cos^{2} (a)) - 3 cos (a) : 2 cos^{2} (a) - 3 cos (a) + 1

2 + cos^{2} (a) - (1 - cos^{2} (a)) - 3 cos (a)

- (1 - cos^{2} (a)) : - 1 + cos^{2} (a)

- (1 - cos^{2} (a))

Poner los parentesis

= - (1) - (- cos^{2} (a))

Aplicar las reglas de los signos

- (- a) = a, - (a) = - a

= - 1 + cos^{2} (a)

= 2 + cos^{2} (a) - 1 + cos^{2} (a) - 3 cos (a)

Simplificar

2 + cos^{2} (a) - 1 + cos^{2} (a) - 3 cos (a) : 2 cos^{2} (a) - 3 cos (a) + 1

2 + cos^{2} (a) - 1 + cos^{2} (a) - 3 cos (a)

Agrupar términos semejantes

= cos^{2} (a) + cos^{2} (a) - 3 cos (a) + 2 - 1

Sumar elementos similares:

cos^{2} (a) + cos^{2} (a) = 2 cos^{2} (a)

= 2 cos^{2} (a) - 3 cos (a) + 2 - 1

Sumar/restar lo siguiente:

2 - 1 = 1

= 2 cos^{2} (a) - 3 cos (a) + 1

= 2 cos^{2} (a) - 3 cos (a) + 1

= 2 cos^{2} (a) - 3 cos (a) + 1

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

Usando el método de sustitución

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

Sea:

cos (a) = u

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

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

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

Escribir en la forma binómica

a x^{2} + b x + c = 0

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

Resolver con la fórmula general para ecuaciones de segundo grado:

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

Formula general para ecuaciones de segundo grado:

Para

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

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

u_{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

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

Separar las soluciones

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

u = \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

u = \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:

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

Sustituir en la ecuación

u = cos (a)

cos (a) = 1, cos (a) = \frac{1}{2}

cos (a) = 1, cos (a) = \frac{1}{2}

cos (a) = 1 : a = 2 πn

cos (a) = 1

Soluciones generales para

cos (a) = 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}

a = 0 + 2 πn

a = 0 + 2 πn

Resolver

a = 0 + 2 πn : a = 2 πn

a = 0 + 2 πn

0 + 2 πn = 2 πn

a = 2 πn

a = 2 πn

cos (a) = \frac{1}{2} : a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn

cos (a) = \frac{1}{2}

Soluciones generales para

cos (a) = \frac{1}{2}

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}

a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn

a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn

Combinar toda las soluciones

a = 2 πn, a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn

Siendo que la ecuación esta indefinida para:

2 πn

a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn

Ejemplos populares

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cos^4(x)=0.37

cos^{4} (x) = 0.37

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sin^{2} (x) = \frac{( 3 )}{2}

sin(x)cos(x)=2sin(2x)

sin (x) cos (x) = 2 sin (2 x)