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1-1/2*cos(3θ)=(4-sqrt(2))/4 ,0<= θ<= 360

Solución

1 - \frac{1}{2} \cdot cos (3 θ) = \frac{4 - 2}{4}, 0^{\circ} \leq θ \leq 36 0^{\circ}

Solución

θ = 1 5^{\circ}, θ = 10 5^{\circ}, θ = 13 5^{\circ}, θ = 22 5^{\circ}, θ = 25 5^{\circ}, θ = 34 5^{\circ}

radianes

θ = \frac{π}{12}, θ = \frac{7 π}{12}, θ = \frac{3 π}{4}, θ = \frac{5 π}{4}, θ = \frac{17 π}{12}, θ = \frac{23 π}{12}

Pasos de solución

1 - \frac{1}{2} cos (3 θ) = \frac{4 - 2}{4}, 0^{\circ} \leq θ \leq 36 0^{\circ}

Mover

1

al lado derecho

1 - \frac{1}{2} cos (3 θ) = \frac{4 - 2}{4}

Restar

1

de ambos lados

1 - \frac{1}{2} cos (3 θ) - 1 = \frac{4 - 2}{4} - 1

Simplificar

- \frac{1}{2} cos (3 θ) = \frac{4 - 2}{4} - 1

- \frac{1}{2} cos (3 θ) = \frac{4 - 2}{4} - 1

Multiplicar ambos lados por

- 2

- \frac{1}{2} cos (3 θ) = \frac{4 - 2}{4} - 1

Multiplicar ambos lados por

- 2

(- \frac{1}{2} cos (3 θ)) (- 2) = \frac{4 - 2}{4} (- 2) - 1 \cdot (- 2)

Simplificar

(- \frac{1}{2} cos (3 θ)) (- 2) = \frac{4 - 2}{4} (- 2) - 1 \cdot (- 2)

Simplificar

(- \frac{1}{2} cos (3 θ)) (- 2) : cos (3 θ)

(- \frac{1}{2} cos (3 θ)) (- 2)

Quitar los parentesis:

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

= \frac{1}{2} cos (3 θ) \cdot 2

Multiplicar fracciones:

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

= \frac{1 \cdot 2}{2} cos (3 θ)

Eliminar los terminos comunes:

2

= cos (3 θ) \cdot 1

Multiplicar:

cos (3 θ) \cdot 1 = cos (3 θ)

= cos (3 θ)

Simplificar

\frac{4 - 2}{4} (- 2) - 1 \cdot (- 2) : - \frac{4 - 2}{2} + 2

\frac{4 - 2}{4} (- 2) - 1 \cdot (- 2)

Quitar los parentesis:

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

= - \frac{4 - 2}{4} \cdot 2 + 1 \cdot 2

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

\frac{4 - 2}{4} \cdot 2

Multiplicar fracciones:

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

= \frac{( 4 - 2 ) \cdot 2}{4}

Eliminar los terminos comunes:

2

= \frac{4 - 2}{2}

1 \cdot 2 = 2

1 \cdot 2

Multiplicar los numeros:

1 \cdot 2 = 2

= 2

= - \frac{4 - 2}{2} + 2

cos (3 θ) = - \frac{4 - 2}{2} + 2

cos (3 θ) = - \frac{4 - 2}{2} + 2

cos (3 θ) = - \frac{4 - 2}{2} + 2

Aplicar propiedades trigonométricas inversas

cos (3 θ) = - \frac{4 - 2}{2} + 2

Soluciones generales para

cos (3 θ) = - \frac{4 - 2}{2} + 2

cos (x) = a \Rightarrow x = arccos (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (a) + 36 0^{\circ} n

3 θ = arccos (- \frac{4 - 2}{2} + 2) + 36 0^{\circ} n, 3 θ = 36 0^{\circ} - arccos (- \frac{4 - 2}{2} + 2) + 36 0^{\circ} n

3 θ = arccos (- \frac{4 - 2}{2} + 2) + 36 0^{\circ} n, 3 θ = 36 0^{\circ} - arccos (- \frac{4 - 2}{2} + 2) + 36 0^{\circ} n

Resolver

3 θ = arccos (- \frac{4 - 2}{2} + 2) + 36 0^{\circ} n : θ = 1 5^{\circ} + \frac{36 0 ^{\circ} n}{3}

3 θ = arccos (- \frac{4 - 2}{2} + 2) + 36 0^{\circ} n

Simplificar

arccos (- \frac{4 - 2}{2} + 2) + 36 0^{\circ} n : 4 5^{\circ} + 36 0^{\circ} n

arccos (- \frac{4 - 2}{2} + 2) + 36 0^{\circ} n

Utilizar la siguiente identidad trivial:

arccos (- \frac{4 - 2}{2} + 2) = 4 5^{\circ}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= 4 5^{\circ} + 36 0^{\circ} n

3 θ = 4 5^{\circ} + 36 0^{\circ} n

Dividir ambos lados entre

3

3 θ = 4 5^{\circ} + 36 0^{\circ} n

Dividir ambos lados entre

3

\frac{3 θ}{3} = \frac{4 5 ^{\circ}}{3} + \frac{36 0 ^{\circ} n}{3}

Simplificar

\frac{3 θ}{3} = \frac{4 5 ^{\circ}}{3} + \frac{36 0 ^{\circ} n}{3}

Simplificar

\frac{3 θ}{3} : θ

\frac{3 θ}{3}

Dividir:

\frac{3}{3} = 1

= θ

Simplificar

\frac{4 5 ^{\circ}}{3} + \frac{36 0 ^{\circ} n}{3} : 1 5^{\circ} + \frac{36 0 ^{\circ} n}{3}

\frac{4 5 ^{\circ}}{3} + \frac{36 0 ^{\circ} n}{3}

\frac{4 5 ^{\circ}}{3} = 1 5^{\circ}

\frac{4 5 ^{\circ}}{3}

Aplicar las propiedades de las fracciones:

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

= \frac{18 0 ^{\circ}}{4 \cdot 3}

Multiplicar los numeros:

4 \cdot 3 = 12

= 1 5^{\circ}

= 1 5^{\circ} + \frac{36 0 ^{\circ} n}{3}

θ = 1 5^{\circ} + \frac{36 0 ^{\circ} n}{3}

θ = 1 5^{\circ} + \frac{36 0 ^{\circ} n}{3}

θ = 1 5^{\circ} + \frac{36 0 ^{\circ} n}{3}

Resolver

3 θ = 36 0^{\circ} - arccos (- \frac{4 - 2}{2} + 2) + 36 0^{\circ} n : θ = 12 0^{\circ} - 1 5^{\circ} + \frac{36 0 ^{\circ} n}{3}

3 θ = 36 0^{\circ} - arccos (- \frac{4 - 2}{2} + 2) + 36 0^{\circ} n

Simplificar

36 0^{\circ} - arccos (- \frac{4 - 2}{2} + 2) + 36 0^{\circ} n : 36 0^{\circ} - 4 5^{\circ} + 36 0^{\circ} n

36 0^{\circ} - arccos (- \frac{4 - 2}{2} + 2) + 36 0^{\circ} n

Utilizar la siguiente identidad trivial:

arccos (- \frac{4 - 2}{2} + 2) = 4 5^{\circ}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= 36 0^{\circ} - 4 5^{\circ} + 36 0^{\circ} n

3 θ = 36 0^{\circ} - 4 5^{\circ} + 36 0^{\circ} n

Dividir ambos lados entre

3

3 θ = 36 0^{\circ} - 4 5^{\circ} + 36 0^{\circ} n

Dividir ambos lados entre

3

\frac{3 θ}{3} = 12 0^{\circ} - \frac{4 5 ^{\circ}}{3} + \frac{36 0 ^{\circ} n}{3}

Simplificar

\frac{3 θ}{3} = 12 0^{\circ} - \frac{4 5 ^{\circ}}{3} + \frac{36 0 ^{\circ} n}{3}

Simplificar

\frac{3 θ}{3} : θ

\frac{3 θ}{3}

Dividir:

\frac{3}{3} = 1

= θ

Simplificar

12 0^{\circ} - \frac{4 5 ^{\circ}}{3} + \frac{36 0 ^{\circ} n}{3} : 12 0^{\circ} - 1 5^{\circ} + \frac{36 0 ^{\circ} n}{3}

12 0^{\circ} - \frac{4 5 ^{\circ}}{3} + \frac{36 0 ^{\circ} n}{3}

\frac{4 5 ^{\circ}}{3} = 1 5^{\circ}

\frac{4 5 ^{\circ}}{3}

Aplicar las propiedades de las fracciones:

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

= \frac{18 0 ^{\circ}}{4 \cdot 3}

Multiplicar los numeros:

4 \cdot 3 = 12

= 1 5^{\circ}

= 12 0^{\circ} - 1 5^{\circ} + \frac{36 0 ^{\circ} n}{3}

θ = 12 0^{\circ} - 1 5^{\circ} + \frac{36 0 ^{\circ} n}{3}

θ = 12 0^{\circ} - 1 5^{\circ} + \frac{36 0 ^{\circ} n}{3}

θ = 12 0^{\circ} - 1 5^{\circ} + \frac{36 0 ^{\circ} n}{3}

θ = 1 5^{\circ} + \frac{36 0 ^{\circ} n}{3}, θ = 12 0^{\circ} - 1 5^{\circ} + \frac{36 0 ^{\circ} n}{3}

Soluciones para el rango

0 \leq θ \leq 36 0^{\circ}

θ = 1 5^{\circ}, θ = 10 5^{\circ}, θ = 13 5^{\circ}, θ = 22 5^{\circ}, θ = 25 5^{\circ}, θ = 34 5^{\circ}

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