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6cos^2(A)-5cos(A)+1=0

Solución

6 cos^{2} (A) - 5 cos (A) + 1 = 0

Solución

A = \frac{π}{3} + 2 πn, A = \frac{5 π}{3} + 2 πn, A = 1.23095 \dots + 2 πn, A = 2 π - 1.23095 \dots + 2 πn

Grados

A = 6 0^{\circ} + 36 0^{\circ} n, A = 30 0^{\circ} + 36 0^{\circ} n, A = 70.52877 \dots^{\circ} + 36 0^{\circ} n, A = 289.47122 \dots^{\circ} + 36 0^{\circ} n

Pasos de solución

6 cos^{2} (A) - 5 cos (A) + 1 = 0

Usando el método de sustitución

6 cos^{2} (A) - 5 cos (A) + 1 = 0

Sea:

cos (A) = u

6 u^{2} - 5 u + 1 = 0

6 u^{2} - 5 u + 1 = 0 : u = \frac{1}{2}, u = \frac{1}{3}

6 u^{2} - 5 u + 1 = 0

Resolver con la fórmula general para ecuaciones de segundo grado:

6 u^{2} - 5 u + 1 = 0

Formula general para ecuaciones de segundo grado:

Para

a = 6, b = - 5, c = 1

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 6 \cdot 1}{2 \cdot 6}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 6 \cdot 1}{2 \cdot 6}

(- 5)^{2} - 4 \cdot 6 \cdot 1 = 1

(- 5)^{2} - 4 \cdot 6 \cdot 1

Aplicar las leyes de los exponentes:

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

n

es par

(- 5)^{2} = 5^{2}

= 5^{2} - 4 \cdot 6 \cdot 1

Multiplicar los numeros:

4 \cdot 6 \cdot 1 = 24

= 5^{2} - 24

5^{2} = 25

= 25 - 24

Restar:

25 - 24 = 1

= 1

Aplicar la regla

1 = 1

= 1

u_{1, 2} = \frac{- ( - 5 ) \pm 1}{2 \cdot 6}

Separar las soluciones

u_{1} = \frac{- ( - 5 ) + 1}{2 \cdot 6}, u_{2} = \frac{- ( - 5 ) - 1}{2 \cdot 6}

u = \frac{- ( - 5 ) + 1}{2 \cdot 6} : \frac{1}{2}

\frac{- ( - 5 ) + 1}{2 \cdot 6}

Aplicar la regla

- (- a) = a

= \frac{5 + 1}{2 \cdot 6}

Sumar:

5 + 1 = 6

= \frac{6}{2 \cdot 6}

Multiplicar los numeros:

2 \cdot 6 = 12

= \frac{6}{12}

Eliminar los terminos comunes:

6

= \frac{1}{2}

u = \frac{- ( - 5 ) - 1}{2 \cdot 6} : \frac{1}{3}

\frac{- ( - 5 ) - 1}{2 \cdot 6}

Aplicar la regla

- (- a) = a

= \frac{5 - 1}{2 \cdot 6}

Restar:

5 - 1 = 4

= \frac{4}{2 \cdot 6}

Multiplicar los numeros:

2 \cdot 6 = 12

= \frac{4}{12}

Eliminar los terminos comunes:

4

= \frac{1}{3}

Las soluciones a la ecuación de segundo grado son:

u = \frac{1}{2}, u = \frac{1}{3}

Sustituir en la ecuación

u = cos (A)

cos (A) = \frac{1}{2}, cos (A) = \frac{1}{3}

cos (A) = \frac{1}{2}, cos (A) = \frac{1}{3}

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

cos (A) = \frac{1}{3} : A = arccos (\frac{1}{3}) + 2 πn, A = 2 π - arccos (\frac{1}{3}) + 2 πn

cos (A) = \frac{1}{3}

Aplicar propiedades trigonométricas inversas

cos (A) = \frac{1}{3}

Soluciones generales para

cos (A) = \frac{1}{3}

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

A = arccos (\frac{1}{3}) + 2 πn, A = 2 π - arccos (\frac{1}{3}) + 2 πn

A = arccos (\frac{1}{3}) + 2 πn, A = 2 π - arccos (\frac{1}{3}) + 2 πn

Combinar toda las soluciones

A = \frac{π}{3} + 2 πn, A = \frac{5 π}{3} + 2 πn, A = arccos (\frac{1}{3}) + 2 πn, A = 2 π - arccos (\frac{1}{3}) + 2 πn

Mostrar soluciones en forma decimal

A = \frac{π}{3} + 2 πn, A = \frac{5 π}{3} + 2 πn, A = 1.23095 \dots + 2 πn, A = 2 π - 1.23095 \dots + 2 πn

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cos(2x-pi/6)=-1/2 , pi/2 <= x<pi

cos (2 x - \frac{π}{6}) = - \frac{1}{2}, \frac{π}{2} \leq x < π

((sin(x)))/((4.2))=((sin(95)))/((10.26))

\frac{( sin ( x ) )}{( 4.2 )} = \frac{( sin ( 9 5 ^{\circ} ) )}{( 10.26 )}