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العربية

-1/2 <cos(x)<= 1/2

Solución

- \frac{1}{2} < cos (x) \leq \frac{1}{2}

\frac{π}{3} + 2 πn \leq x < \frac{2 π}{3} + 2 πn or \frac{4 π}{3} + 2 πn < x \leq \frac{5 π}{3} + 2 πn

Notación de intervalos

[\frac{π}{3} + 2 πn, \frac{2 π}{3} + 2 πn) \cup (\frac{4 π}{3} + 2 πn, \frac{5 π}{3} + 2 πn]

Decimal

1.04719 \dots + 2 πn \leq x < 2.09439 \dots + 2 πn or 4.18879 \dots + 2 πn < x \leq 5.23598 \dots + 2 πn

Pasos de solución

- \frac{1}{2} < cos (x) \leq \frac{1}{2}

a < u \leq b

entonces

a < u and u \leq b

- \frac{1}{2} < cos (x) and cos (x) \leq \frac{1}{2}

- \frac{1}{2} < cos (x) : - \frac{2 π}{3} + 2 πn < x < \frac{2 π}{3} + 2 πn

- \frac{1}{2} < cos (x)

Intercambiar lados

cos (x) > - \frac{1}{2}

Para

cos (x) > a

, si

- 1 \leq a < 1

entonces

- arccos (a) + 2 πn < x < arccos (a) + 2 πn

- arccos (- \frac{1}{2}) + 2 πn < x < arccos (- \frac{1}{2}) + 2 πn

Simplificar

- arccos (- \frac{1}{2}) : - \frac{2 π}{3}

- arccos (- \frac{1}{2})

Utilizar la siguiente identidad trivial:

arccos (- \frac{1}{2}) = \frac{2 π}{3}

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}

= - \frac{2 π}{3}

Simplificar

arccos (- \frac{1}{2}) : \frac{2 π}{3}

arccos (- \frac{1}{2})

Utilizar la siguiente identidad trivial:

arccos (- \frac{1}{2}) = \frac{2 π}{3}

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}

= \frac{2 π}{3}

- \frac{2 π}{3} + 2 πn < x < \frac{2 π}{3} + 2 πn

cos (x) \leq \frac{1}{2} : \frac{π}{3} + 2 πn \leq x \leq \frac{5 π}{3} + 2 πn

cos (x) \leq \frac{1}{2}

Para

cos (x) \leq a

, si

- 1 < a < 1

entonces

arccos (a) + 2 πn \leq x \leq 2 π - arccos (a) + 2 πn

arccos (\frac{1}{2}) + 2 πn \leq x \leq 2 π - arccos (\frac{1}{2}) + 2 πn

Simplificar

arccos (\frac{1}{2}) : \frac{π}{3}

arccos (\frac{1}{2})

Utilizar la siguiente identidad trivial:

arccos (\frac{1}{2}) = \frac{π}{3}

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}

= \frac{π}{3}

Simplificar

2 π - arccos (\frac{1}{2}) : \frac{5 π}{3}

2 π - arccos (\frac{1}{2})

Utilizar la siguiente identidad trivial:

arccos (\frac{1}{2}) = \frac{π}{3}

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}

= 2 π - \frac{π}{3}

Simplificar

2 π - \frac{π}{3}

Convertir a fracción:

2 π = \frac{2 π 3}{3}

= \frac{2 π 3}{3} - \frac{π}{3}

Ya que los denominadores son iguales, combinar las fracciones:

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

= \frac{2 π 3 - π}{3}

2 π 3 - π = 5 π

2 π 3 - π

Multiplicar los numeros:

2 \cdot 3 = 6

= 6 π - π

Sumar elementos similares:

6 π - π = 5 π

= 5 π

= \frac{5 π}{3}

= \frac{5 π}{3}

\frac{π}{3} + 2 πn \leq x \leq \frac{5 π}{3} + 2 πn

Combinar los rangos

- \frac{2 π}{3} + 2 πn < x < \frac{2 π}{3} + 2 πn and \frac{π}{3} + 2 πn \leq x \leq \frac{5 π}{3} + 2 πn

Mezclar intervalos sobrepuestos

\frac{π}{3} + 2 πn \leq x < \frac{2 π}{3} + 2 πn or \frac{4 π}{3} + 2 πn < x \leq \frac{5 π}{3} + 2 πn