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cos^2(x)>= 1/2

Solución

cos^{2} (x) \geq \frac{1}{2}

Solución

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

Notación de intervalos

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

Notación decimal

- 0.78539 \dots + 2 πn \leq x \leq 0.78539 \dots + 2 πn or 2.35619 \dots + 2 πn \leq x \leq 3.92699 \dots + 2 πn

Pasos de solución

cos^{2} (x) \geq \frac{1}{2}

Para

u^{n} \geq a

, si

n

es par entonces

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

cos (x) \leq - \frac{1}{2} : \frac{3 π}{4} + 2 πn \leq x \leq \frac{5 π}{4} + 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{3 π}{4}

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

Utilizar la siguiente identidad trivial:

arccos (- \frac{1}{2}) = \frac{3 π}{4}

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{3 π}{4}

Simplificar

2 π - arccos (- \frac{1}{2}) : \frac{5 π}{4}

2 π - arccos (- \frac{1}{2})

Utilizar la siguiente identidad trivial:

arccos (- \frac{1}{2}) = \frac{3 π}{4}

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{3 π}{4}

Simplificar

2 π - \frac{3 π}{4}

Convertir a fracción:

2 π = \frac{2 π 4}{4}

= \frac{2 π 4}{4} - \frac{3 π}{4}

Ya que los denominadores son iguales, combinar las fracciones:

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

= \frac{2 π 4 - 3 π}{4}

2 π 4 - 3 π = 5 π

2 π 4 - 3 π

Multiplicar los numeros:

2 \cdot 4 = 8

= 8 π - 3 π

Sumar elementos similares:

8 π - 3 π = 5 π

= 5 π

= \frac{5 π}{4}

= \frac{5 π}{4}

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

cos (x) \geq \frac{1}{2} : - \frac{π}{4} + 2 πn \leq x \leq \frac{π}{4} + 2 πn

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

Para

cos (x) \geq a

, si

- 1 < a < 1

entonces

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

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

Simplificar

- arccos (\frac{1}{2}) : - \frac{π}{4}

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

Utilizar la siguiente identidad trivial:

arccos (\frac{1}{2}) = \frac{π}{4}

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}{4}

Simplificar

arccos (\frac{1}{2}) : \frac{π}{4}

arccos (\frac{1}{2})

Utilizar la siguiente identidad trivial:

arccos (\frac{1}{2}) = \frac{π}{4}

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}{4}

- \frac{π}{4} + 2 πn \leq x \leq \frac{π}{4} + 2 πn

Combinar los rangos

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

Mezclar intervalos sobrepuestos

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

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