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2cos^2(x)<1

Solución

2 cos^{2} (x) < 1

Solución

\frac{π}{4} + 2 πn < x < \frac{3 π}{4} + 2 πn or \frac{5 π}{4} + 2 πn < x < \frac{7 π}{4} + 2 πn

Notación de intervalos

(\frac{π}{4} + 2 πn, \frac{3 π}{4} + 2 πn) \cup (\frac{5 π}{4} + 2 πn, \frac{7 π}{4} + 2 πn)

Notación decimal

0.78539 \dots + 2 πn < x < 2.35619 \dots + 2 πn or 3.92699 \dots + 2 πn < x < 5.49778 \dots + 2 πn

Pasos de solución

2 cos^{2} (x) < 1

Dividir ambos lados entre

2

2 cos^{2} (x) < 1

Dividir ambos lados entre

2

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

Simplificar

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

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

Para

u^{n} < a

, si

n

es par entonces

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

a < u < b

entonces

a < u and u < b

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

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

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}

- \frac{3 π}{4} + 2 πn < x < \frac{3 π}{4} + 2 πn

cos (x) < \frac{1}{2} : \frac{π}{4} + 2 πn < x < \frac{7 π}{4} + 2 πn

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

Para

cos (x) < a

, si

- 1 < a \leq 1

entonces

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

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

2 π - arccos (\frac{1}{2}) : \frac{7 π}{4}

2 π - 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}

= 2 π - \frac{π}{4}

Simplificar

2 π - \frac{π}{4}

Convertir a fracción:

2 π = \frac{2 π 4}{4}

= \frac{2 π 4}{4} - \frac{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 - π}{4}

2 π 4 - π = 7 π

2 π 4 - π

Multiplicar los numeros:

2 \cdot 4 = 8

= 8 π - π

Sumar elementos similares:

8 π - π = 7 π

= 7 π

= \frac{7 π}{4}

= \frac{7 π}{4}

\frac{π}{4} + 2 πn < x < \frac{7 π}{4} + 2 πn

Combinar los rangos

- \frac{3 π}{4} + 2 πn < x < \frac{3 π}{4} + 2 πn and \frac{π}{4} + 2 πn < x < \frac{7 π}{4} + 2 πn

Mezclar intervalos sobrepuestos

\frac{π}{4} + 2 πn < x < \frac{3 π}{4} + 2 πn or \frac{5 π}{4} + 2 πn < x < \frac{7 π}{4} + 2 πn

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